Counting,Proability and Binomial Coefficients
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If $$P_{2n+2}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^n-{2n choose {n+1}}p^{n+1}q^{n+1}$$
$mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
probability sequences-and-series combinatorics probability-theory binomial-coefficients
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
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If $$P_{2n+2}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^n-{2n choose {n+1}}p^{n+1}q^{n+1}$$
$mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
probability sequences-and-series combinatorics probability-theory binomial-coefficients
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
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This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
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– ablmf
Jan 13 at 10:30
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If $$P_{2n+2}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^n-{2n choose {n+1}}p^{n+1}q^{n+1}$$
$mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
probability sequences-and-series combinatorics probability-theory binomial-coefficients
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
$endgroup$
If $$P_{2n+2}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^n-{2n choose {n+1}}p^{n+1}q^{n+1}$$
$mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
probability sequences-and-series combinatorics probability-theory binomial-coefficients
probability sequences-and-series combinatorics probability-theory binomial-coefficients
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
asked Jan 13 at 4:28
Arijit DeyArijit Dey
314
314
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This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
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– ablmf
Jan 13 at 10:30
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This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
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– ablmf
Jan 13 at 10:30
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This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
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– ablmf
Jan 13 at 10:30
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This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
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– ablmf
Jan 13 at 10:30
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4 Answers
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We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p^2+2pq+q^2)=sum_{k=n+1}^{2n}{2n choose k}p^{k+2}q^{2n-k}+sum_{k=n+1}^{2n}2{2n choose k}p^{k+1}q^{2n-k+1}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}=sum_{k=n+3}^{2n+2}{2n choose k-2}p^{k}q^{2n+2-k}+sum_{k=n+2}^{2n+1}2{2n choose k-1}p^{k}q^{2n+2-k}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
sum_{k=n+2}^{2n+2}left({2n choose k}+2{2n choose k-1}+{2n choose k-2}right)p^kq^{2n+2-k}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
We also know that
$$
{2n choose k}+2{2n choose k-1}+{2n choose k-2}=left({2n choose k}+{2n choose k-1}right)+left({2n choose k-1}+{2n choose k-2}right)={2n+1 choose k}+{2n+1 choose k-1}={2n+2 choose k},
$$
so we obtain
$$
P_{2n}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}=P_{2n+2}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1},
$$
as desired.
answered Jan 13 at 13:10
richrowrichrow
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The equality reduces to
$$
binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} , _2F_1left(1,1-n;n+2;frac{p}{p-1}right)+binom{2 n}{n} p^{n+2} (1-p)^n-binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)
$$
Let $w=p/(1-p)$, then the above is
$$
binom{2 n}{n+1} left((w+1)^2 , _2F_1(1,1-n;n+2;-w)-1right)+w binom{2 n}{n}=w binom{2 (n+1)}{n+2} , _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m in {0,dots,n}$ to see that this equality holds.
answered Jan 13 at 11:23
ablmfablmf
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Forget the requirement that $0<p<q<1$; it is unnecessary. So let $p$ and $q$
be any reals satisfying $p+q=1$.
Also, let me extend the stage a little bit:
Definition. Let $m$ and $n$ be integers such that $ngeq mgeq0$. Then, we
let
begin{equation}
Q_{n,m}=sum_{k=m}^{n}dbinom{n}{k}p^{k}q^{n-k}.
end{equation}
With this definition, your $P_{2n}$ is $Q_{2n,n+1}$, while your $P_{2n+2}$ is
$Q_{2n+2,n+2}$. Thus, your claim becomes:
Theorem 1. Let $n$ be a nonnegative integer. Then,
begin{equation}
Q_{2n+2,n+2}=Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}
p^{n+1}q^{n+1}.
end{equation}
I shall derive this from the following:
Lemma 2. Let $m$ and $n$ be integers such that $ngeq mgeq1$. Then,
begin{equation}
Q_{n,m}=Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{equation}
Proof of Lemma 2. From $ngeq mgeq1$, we obtain $n-1geq m-1geq0$. Now,
the definition of $Q_{n-1,m-1}$ yields
begin{equation}
Q_{n-1,m-1}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}underbrace{q^{n-1-k}
}_{=q^{n-k-1}}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}.
label{darij1.pf.l2.0}
tag{1}
end{equation}
But the definition of $Q_{n,m}$ yields
begin{align}
Q_{n,m} & =sum_{k=m}^{n}underbrace{dbinom{n}{k}}_{substack{=dbinom
{n-1}{k-1}+dbinom{n-1}{k}\text{(by the recurrence of the binomial
coefficients)}}}p^{k}q^{n-k}=sum_{k=m}^{n}left( dbinom{n-1}{k-1}
+dbinom{n-1}{k}right) p^{k}q^{n-k}nonumber\
& =sum_{k=m}^{n}dbinom{n-1}{k-1}p^{k}q^{n-k}+sum_{k=m}^{n}dbinom{n-1}
{k}p^{k}q^{n-k}nonumber\
& =sum_{k=m-1}^{n-1}underbrace{dbinom{n-1}{left( k+1right) -1}
}_{=dbinom{n-1}{k}}p^{k+1}underbrace{q^{n-left( k+1right) }}
_{=q^{n-k-1}}+sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}nonumber\
& qquadleft( text{here, we have substituted }k+1text{ for }ktext{ in
the first sum}right) nonumber\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum_{k=m}^{n}
dbinom{n-1}{k}p^{k}q^{n-k}.
label{darij1.pf.l2.1}
tag{2}
end{align}
But we have
begin{equation}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m}^{n}dbinom{n-1}{k}
p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }
end{equation}
and
begin{align*}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k} & =sum_{k=m-1}^{n-1}dbinom
{n-1}{k}p^{k}q^{n-k}+underbrace{dbinom{n-1}{n}}_{substack{=0\text{(since
}n-1geq0text{ and }n-1<ntext{)}}}p^{n}q^{n-n}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{align*}
Comparing these two equalities, we obtain
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left(
m-1right) }=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{equation}
Thus,
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }.
end{equation}
Hence, eqref{darij1.pf.l2.1} becomes
begin{align*}
Q_{n,m} & =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+underbrace{sum
_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum
_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}left( p^{k+1}q^{n-k-1}+p^{k}q^{n-k}right) }-dbinom{n-1}{m-1}
p^{m-1}underbrace{q^{n-left( m-1right) }}_{=q^{n-m+1}}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}left( underbrace{p^{k+1}}_{=pp^{k}
}q^{n-k-1}+p^{k}underbrace{q^{n-k}}_{=qq^{n-k-1}}right) -dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}underbrace{left( pp^{k}q^{n-k-1}
+p^{k}qq^{n-k-1}right) }_{substack{=left( p+qright) p^{k}
q^{n-k-1}=p^{k}q^{n-k-1}\text{(since }p+q=1text{)}}}-dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}}
_{substack{=Q_{n-1,m-1}\text{(by eqref{darij1.pf.l2.0})}}}-dbinom
{n-1}{m-1}p^{m-1}q^{n-m+1}\
& =Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{align*}
This proves Lemma 2. $blacksquare$
Proof of Theorem 1. We have $left( 1-q-pqright) -p^{2}=-left(
p+1right) underbrace{left( p+q-1right) }_{substack{=0\text{(since
}p+q=1text{)}}}=0$, thus $1-q-pq=p^{2}$.
Lemma 2 (applied to $2n+2$ and $n+2$ instead of $n$ and $m$) yields
begin{align}
Q_{2n+2,n+2} & =underbrace{Q_{2n+1,n+1}}_{substack{=Q_{2n,n}-dbinom{2n}
{n}p^{n}q^{left( 2n+1right) -left( n+1right) +1}\text{(by Lemma 2,
applied to }2n+1text{ and }n+1\text{instead of }ntext{ and }mtext{)}
}}-underbrace{dbinom{2n+1}{n+1}}_{substack{=dbinom{2n}{n}+dbinom{2n}
{n+1}\text{(by the recurrence of the binomial coefficients)}}}p^{n+1}
underbrace{q^{left( 2n+2right) -left( n+2right) +1}}_{=q^{n+1}
}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}underbrace{q^{left( 2n+1right) -left(
n+1right) +1}}_{=q^{n+1}}-underbrace{left( dbinom{2n}{n}+dbinom
{2n}{n+1}right) p^{n+1}q^{n+1}}_{=dbinom{2n}{n}p^{n+1}q^{n+1}+dbinom
{2n}{n+1}p^{n+1}q^{n+1}}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-left( dbinom{2n}{n}p^{n+1}
q^{n+1}+dbinom{2n}{n+1}p^{n+1}q^{n+1}right) nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
label{darij1.pf.t1.1}
tag{3}
end{align}
But the definition of $Q_{2n,n+1}$ yields $Q_{2n,n+1}=sum_{k=n+1}^{2n}
dbinom{2n}{k}p^{k}q^{2n-k}$. Meanwhile, the definition of $Q_{2n,n}$ yields
begin{align*}
Q_{2n,n} & =sum_{k=n}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}=underbrace{sum
_{k=n+1}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}}_{=Q_{2n,n+1}}+dbinom{2n}{n}
p^{n}underbrace{q^{2n-n}}_{=q^{n}}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}.
end{align*}
Thus,
eqref{darij1.pf.t1.1} becomes
begin{align*}
Q_{2n+2,n+2} & =underbrace{Q_{2n,n}}_{=Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}
}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}
{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom
{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( p^{n}q^{n}-p^{n}
q^{n+1}-p^{n+1}q^{n+1}right) }_{=left( 1-q-pqright) p^{n}q^{n}}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( 1-q-pqright) }_{=p^{2}}
p^{n}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{p^{2}p^{n}}_{=p^{n+2}}q^{n}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
end{align*}
This proves Theorem 1. $blacksquare$
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
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We have for the first sum
$$S = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = sum_{k=0}^{n-1} {2nchoose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
sum_{k=0}^n {2n+2choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
sum_{k=0}^n {2n+2choose n-k} p^{k} q^{-k}
\ = p^{n+2} q^n
sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{kge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} frac{1}{1-pz/q}
\ = p^2 [z^n] (1+pqz)^{2n+2} frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] z (1+pqz)^{2n} frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\ = p^2 p^n q^n {2nchoose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\ = p^{n+2} q^n {2nchoose n}
- p^2 q^2 p^{n-1} q^{n-1} {2nchoose n-1}.$$
This is indeed
$$bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2nchoose n}
- p^{n+1} q^{n+1} {2nchoose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
add a comment |
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$begingroup$
We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p^2+2pq+q^2)=sum_{k=n+1}^{2n}{2n choose k}p^{k+2}q^{2n-k}+sum_{k=n+1}^{2n}2{2n choose k}p^{k+1}q^{2n-k+1}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}=sum_{k=n+3}^{2n+2}{2n choose k-2}p^{k}q^{2n+2-k}+sum_{k=n+2}^{2n+1}2{2n choose k-1}p^{k}q^{2n+2-k}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
sum_{k=n+2}^{2n+2}left({2n choose k}+2{2n choose k-1}+{2n choose k-2}right)p^kq^{2n+2-k}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
We also know that
$$
{2n choose k}+2{2n choose k-1}+{2n choose k-2}=left({2n choose k}+{2n choose k-1}right)+left({2n choose k-1}+{2n choose k-2}right)={2n+1 choose k}+{2n+1 choose k-1}={2n+2 choose k},
$$
so we obtain
$$
P_{2n}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}=P_{2n+2}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1},
$$
as desired.
answered Jan 13 at 13:10
richrowrichrow
1615
$endgroup$
add a comment |
$begingroup$
We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p^2+2pq+q^2)=sum_{k=n+1}^{2n}{2n choose k}p^{k+2}q^{2n-k}+sum_{k=n+1}^{2n}2{2n choose k}p^{k+1}q^{2n-k+1}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}=sum_{k=n+3}^{2n+2}{2n choose k-2}p^{k}q^{2n+2-k}+sum_{k=n+2}^{2n+1}2{2n choose k-1}p^{k}q^{2n+2-k}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
sum_{k=n+2}^{2n+2}left({2n choose k}+2{2n choose k-1}+{2n choose k-2}right)p^kq^{2n+2-k}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
We also know that
$$
{2n choose k}+2{2n choose k-1}+{2n choose k-2}=left({2n choose k}+{2n choose k-1}right)+left({2n choose k-1}+{2n choose k-2}right)={2n+1 choose k}+{2n+1 choose k-1}={2n+2 choose k},
$$
so we obtain
$$
P_{2n}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}=P_{2n+2}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1},
$$
as desired.
answered Jan 13 at 13:10
richrowrichrow
1615
$endgroup$
add a comment |
$begingroup$
We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p^2+2pq+q^2)=sum_{k=n+1}^{2n}{2n choose k}p^{k+2}q^{2n-k}+sum_{k=n+1}^{2n}2{2n choose k}p^{k+1}q^{2n-k+1}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}=sum_{k=n+3}^{2n+2}{2n choose k-2}p^{k}q^{2n+2-k}+sum_{k=n+2}^{2n+1}2{2n choose k-1}p^{k}q^{2n+2-k}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
sum_{k=n+2}^{2n+2}left({2n choose k}+2{2n choose k-1}+{2n choose k-2}right)p^kq^{2n+2-k}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
We also know that
$$
{2n choose k}+2{2n choose k-1}+{2n choose k-2}=left({2n choose k}+{2n choose k-1}right)+left({2n choose k-1}+{2n choose k-2}right)={2n+1 choose k}+{2n+1 choose k-1}={2n+2 choose k},
$$
so we obtain
$$
P_{2n}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}=P_{2n+2}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1},
$$
as desired.
answered Jan 13 at 13:10
richrowrichrow
1615
$endgroup$
We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k}=sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
sum_{k=n+1}^{2n}{2n choose k}p^kq^{2n-k} (p^2+2pq+q^2)=sum_{k=n+1}^{2n}{2n choose k}p^{k+2}q^{2n-k}+sum_{k=n+1}^{2n}2{2n choose k}p^{k+1}q^{2n-k+1}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}=sum_{k=n+3}^{2n+2}{2n choose k-2}p^{k}q^{2n+2-k}+sum_{k=n+2}^{2n+1}2{2n choose k-1}p^{k}q^{2n+2-k}+sum_{k=n+1}^{2n}{2n choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
sum_{k=n+2}^{2n+2}left({2n choose k}+2{2n choose k-1}+{2n choose k-2}right)p^kq^{2n+2-k}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
We also know that
$$
{2n choose k}+2{2n choose k-1}+{2n choose k-2}=left({2n choose k}+{2n choose k-1}right)+left({2n choose k-1}+{2n choose k-2}right)={2n+1 choose k}+{2n+1 choose k-1}={2n+2 choose k},
$$
so we obtain
$$
P_{2n}=sum_{k=n+2}^{2n+2}{2n+2 choose k}p^kq^{2n+2-k}=P_{2n+2}-left({2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1}right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n choose n}p^{n+2}q^{n}-{2n choose n+1}p^{n+1}q^{n+1},
$$
as desired.
answered Jan 13 at 13:10
richrowrichrow
1615
answered Jan 13 at 13:10
richrowrichrow
1615
answered Jan 13 at 13:10
richrowrichrow
1615
answered Jan 13 at 13:10
richrowrichrow
1615
1615
add a comment |
add a comment |
$begingroup$
The equality reduces to
$$
binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} , _2F_1left(1,1-n;n+2;frac{p}{p-1}right)+binom{2 n}{n} p^{n+2} (1-p)^n-binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)
$$
Let $w=p/(1-p)$, then the above is
$$
binom{2 n}{n+1} left((w+1)^2 , _2F_1(1,1-n;n+2;-w)-1right)+w binom{2 n}{n}=w binom{2 (n+1)}{n+2} , _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m in {0,dots,n}$ to see that this equality holds.
answered Jan 13 at 11:23
ablmfablmf
2,52142352
$endgroup$
add a comment |
$begingroup$
The equality reduces to
$$
binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} , _2F_1left(1,1-n;n+2;frac{p}{p-1}right)+binom{2 n}{n} p^{n+2} (1-p)^n-binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)
$$
Let $w=p/(1-p)$, then the above is
$$
binom{2 n}{n+1} left((w+1)^2 , _2F_1(1,1-n;n+2;-w)-1right)+w binom{2 n}{n}=w binom{2 (n+1)}{n+2} , _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m in {0,dots,n}$ to see that this equality holds.
answered Jan 13 at 11:23
ablmfablmf
2,52142352
$endgroup$
add a comment |
$begingroup$
The equality reduces to
$$
binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} , _2F_1left(1,1-n;n+2;frac{p}{p-1}right)+binom{2 n}{n} p^{n+2} (1-p)^n-binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)
$$
Let $w=p/(1-p)$, then the above is
$$
binom{2 n}{n+1} left((w+1)^2 , _2F_1(1,1-n;n+2;-w)-1right)+w binom{2 n}{n}=w binom{2 (n+1)}{n+2} , _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m in {0,dots,n}$ to see that this equality holds.
answered Jan 13 at 11:23
ablmfablmf
2,52142352
$endgroup$
The equality reduces to
$$
binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} , _2F_1left(1,1-n;n+2;frac{p}{p-1}right)+binom{2 n}{n} p^{n+2} (1-p)^n-binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)
$$
Let $w=p/(1-p)$, then the above is
$$
binom{2 n}{n+1} left((w+1)^2 , _2F_1(1,1-n;n+2;-w)-1right)+w binom{2 n}{n}=w binom{2 (n+1)}{n+2} , _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m in {0,dots,n}$ to see that this equality holds.
answered Jan 13 at 11:23
ablmfablmf
2,52142352
answered Jan 13 at 11:23
ablmfablmf
2,52142352
answered Jan 13 at 11:23
ablmfablmf
2,52142352
answered Jan 13 at 11:23
ablmfablmf
2,52142352
2,52142352
add a comment |
add a comment |
$begingroup$
Forget the requirement that $0<p<q<1$; it is unnecessary. So let $p$ and $q$
be any reals satisfying $p+q=1$.
Also, let me extend the stage a little bit:
Definition. Let $m$ and $n$ be integers such that $ngeq mgeq0$. Then, we
let
begin{equation}
Q_{n,m}=sum_{k=m}^{n}dbinom{n}{k}p^{k}q^{n-k}.
end{equation}
With this definition, your $P_{2n}$ is $Q_{2n,n+1}$, while your $P_{2n+2}$ is
$Q_{2n+2,n+2}$. Thus, your claim becomes:
Theorem 1. Let $n$ be a nonnegative integer. Then,
begin{equation}
Q_{2n+2,n+2}=Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}
p^{n+1}q^{n+1}.
end{equation}
I shall derive this from the following:
Lemma 2. Let $m$ and $n$ be integers such that $ngeq mgeq1$. Then,
begin{equation}
Q_{n,m}=Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{equation}
Proof of Lemma 2. From $ngeq mgeq1$, we obtain $n-1geq m-1geq0$. Now,
the definition of $Q_{n-1,m-1}$ yields
begin{equation}
Q_{n-1,m-1}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}underbrace{q^{n-1-k}
}_{=q^{n-k-1}}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}.
label{darij1.pf.l2.0}
tag{1}
end{equation}
But the definition of $Q_{n,m}$ yields
begin{align}
Q_{n,m} & =sum_{k=m}^{n}underbrace{dbinom{n}{k}}_{substack{=dbinom
{n-1}{k-1}+dbinom{n-1}{k}\text{(by the recurrence of the binomial
coefficients)}}}p^{k}q^{n-k}=sum_{k=m}^{n}left( dbinom{n-1}{k-1}
+dbinom{n-1}{k}right) p^{k}q^{n-k}nonumber\
& =sum_{k=m}^{n}dbinom{n-1}{k-1}p^{k}q^{n-k}+sum_{k=m}^{n}dbinom{n-1}
{k}p^{k}q^{n-k}nonumber\
& =sum_{k=m-1}^{n-1}underbrace{dbinom{n-1}{left( k+1right) -1}
}_{=dbinom{n-1}{k}}p^{k+1}underbrace{q^{n-left( k+1right) }}
_{=q^{n-k-1}}+sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}nonumber\
& qquadleft( text{here, we have substituted }k+1text{ for }ktext{ in
the first sum}right) nonumber\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum_{k=m}^{n}
dbinom{n-1}{k}p^{k}q^{n-k}.
label{darij1.pf.l2.1}
tag{2}
end{align}
But we have
begin{equation}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m}^{n}dbinom{n-1}{k}
p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }
end{equation}
and
begin{align*}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k} & =sum_{k=m-1}^{n-1}dbinom
{n-1}{k}p^{k}q^{n-k}+underbrace{dbinom{n-1}{n}}_{substack{=0\text{(since
}n-1geq0text{ and }n-1<ntext{)}}}p^{n}q^{n-n}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{align*}
Comparing these two equalities, we obtain
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left(
m-1right) }=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{equation}
Thus,
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }.
end{equation}
Hence, eqref{darij1.pf.l2.1} becomes
begin{align*}
Q_{n,m} & =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+underbrace{sum
_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum
_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}left( p^{k+1}q^{n-k-1}+p^{k}q^{n-k}right) }-dbinom{n-1}{m-1}
p^{m-1}underbrace{q^{n-left( m-1right) }}_{=q^{n-m+1}}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}left( underbrace{p^{k+1}}_{=pp^{k}
}q^{n-k-1}+p^{k}underbrace{q^{n-k}}_{=qq^{n-k-1}}right) -dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}underbrace{left( pp^{k}q^{n-k-1}
+p^{k}qq^{n-k-1}right) }_{substack{=left( p+qright) p^{k}
q^{n-k-1}=p^{k}q^{n-k-1}\text{(since }p+q=1text{)}}}-dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}}
_{substack{=Q_{n-1,m-1}\text{(by eqref{darij1.pf.l2.0})}}}-dbinom
{n-1}{m-1}p^{m-1}q^{n-m+1}\
& =Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{align*}
This proves Lemma 2. $blacksquare$
Proof of Theorem 1. We have $left( 1-q-pqright) -p^{2}=-left(
p+1right) underbrace{left( p+q-1right) }_{substack{=0\text{(since
}p+q=1text{)}}}=0$, thus $1-q-pq=p^{2}$.
Lemma 2 (applied to $2n+2$ and $n+2$ instead of $n$ and $m$) yields
begin{align}
Q_{2n+2,n+2} & =underbrace{Q_{2n+1,n+1}}_{substack{=Q_{2n,n}-dbinom{2n}
{n}p^{n}q^{left( 2n+1right) -left( n+1right) +1}\text{(by Lemma 2,
applied to }2n+1text{ and }n+1\text{instead of }ntext{ and }mtext{)}
}}-underbrace{dbinom{2n+1}{n+1}}_{substack{=dbinom{2n}{n}+dbinom{2n}
{n+1}\text{(by the recurrence of the binomial coefficients)}}}p^{n+1}
underbrace{q^{left( 2n+2right) -left( n+2right) +1}}_{=q^{n+1}
}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}underbrace{q^{left( 2n+1right) -left(
n+1right) +1}}_{=q^{n+1}}-underbrace{left( dbinom{2n}{n}+dbinom
{2n}{n+1}right) p^{n+1}q^{n+1}}_{=dbinom{2n}{n}p^{n+1}q^{n+1}+dbinom
{2n}{n+1}p^{n+1}q^{n+1}}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-left( dbinom{2n}{n}p^{n+1}
q^{n+1}+dbinom{2n}{n+1}p^{n+1}q^{n+1}right) nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
label{darij1.pf.t1.1}
tag{3}
end{align}
But the definition of $Q_{2n,n+1}$ yields $Q_{2n,n+1}=sum_{k=n+1}^{2n}
dbinom{2n}{k}p^{k}q^{2n-k}$. Meanwhile, the definition of $Q_{2n,n}$ yields
begin{align*}
Q_{2n,n} & =sum_{k=n}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}=underbrace{sum
_{k=n+1}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}}_{=Q_{2n,n+1}}+dbinom{2n}{n}
p^{n}underbrace{q^{2n-n}}_{=q^{n}}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}.
end{align*}
Thus,
eqref{darij1.pf.t1.1} becomes
begin{align*}
Q_{2n+2,n+2} & =underbrace{Q_{2n,n}}_{=Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}
}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}
{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom
{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( p^{n}q^{n}-p^{n}
q^{n+1}-p^{n+1}q^{n+1}right) }_{=left( 1-q-pqright) p^{n}q^{n}}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( 1-q-pqright) }_{=p^{2}}
p^{n}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{p^{2}p^{n}}_{=p^{n+2}}q^{n}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
end{align*}
This proves Theorem 1. $blacksquare$
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
$endgroup$
add a comment |
$begingroup$
Forget the requirement that $0<p<q<1$; it is unnecessary. So let $p$ and $q$
be any reals satisfying $p+q=1$.
Also, let me extend the stage a little bit:
Definition. Let $m$ and $n$ be integers such that $ngeq mgeq0$. Then, we
let
begin{equation}
Q_{n,m}=sum_{k=m}^{n}dbinom{n}{k}p^{k}q^{n-k}.
end{equation}
With this definition, your $P_{2n}$ is $Q_{2n,n+1}$, while your $P_{2n+2}$ is
$Q_{2n+2,n+2}$. Thus, your claim becomes:
Theorem 1. Let $n$ be a nonnegative integer. Then,
begin{equation}
Q_{2n+2,n+2}=Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}
p^{n+1}q^{n+1}.
end{equation}
I shall derive this from the following:
Lemma 2. Let $m$ and $n$ be integers such that $ngeq mgeq1$. Then,
begin{equation}
Q_{n,m}=Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{equation}
Proof of Lemma 2. From $ngeq mgeq1$, we obtain $n-1geq m-1geq0$. Now,
the definition of $Q_{n-1,m-1}$ yields
begin{equation}
Q_{n-1,m-1}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}underbrace{q^{n-1-k}
}_{=q^{n-k-1}}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}.
label{darij1.pf.l2.0}
tag{1}
end{equation}
But the definition of $Q_{n,m}$ yields
begin{align}
Q_{n,m} & =sum_{k=m}^{n}underbrace{dbinom{n}{k}}_{substack{=dbinom
{n-1}{k-1}+dbinom{n-1}{k}\text{(by the recurrence of the binomial
coefficients)}}}p^{k}q^{n-k}=sum_{k=m}^{n}left( dbinom{n-1}{k-1}
+dbinom{n-1}{k}right) p^{k}q^{n-k}nonumber\
& =sum_{k=m}^{n}dbinom{n-1}{k-1}p^{k}q^{n-k}+sum_{k=m}^{n}dbinom{n-1}
{k}p^{k}q^{n-k}nonumber\
& =sum_{k=m-1}^{n-1}underbrace{dbinom{n-1}{left( k+1right) -1}
}_{=dbinom{n-1}{k}}p^{k+1}underbrace{q^{n-left( k+1right) }}
_{=q^{n-k-1}}+sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}nonumber\
& qquadleft( text{here, we have substituted }k+1text{ for }ktext{ in
the first sum}right) nonumber\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum_{k=m}^{n}
dbinom{n-1}{k}p^{k}q^{n-k}.
label{darij1.pf.l2.1}
tag{2}
end{align}
But we have
begin{equation}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m}^{n}dbinom{n-1}{k}
p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }
end{equation}
and
begin{align*}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k} & =sum_{k=m-1}^{n-1}dbinom
{n-1}{k}p^{k}q^{n-k}+underbrace{dbinom{n-1}{n}}_{substack{=0\text{(since
}n-1geq0text{ and }n-1<ntext{)}}}p^{n}q^{n-n}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{align*}
Comparing these two equalities, we obtain
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left(
m-1right) }=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{equation}
Thus,
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }.
end{equation}
Hence, eqref{darij1.pf.l2.1} becomes
begin{align*}
Q_{n,m} & =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+underbrace{sum
_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum
_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}left( p^{k+1}q^{n-k-1}+p^{k}q^{n-k}right) }-dbinom{n-1}{m-1}
p^{m-1}underbrace{q^{n-left( m-1right) }}_{=q^{n-m+1}}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}left( underbrace{p^{k+1}}_{=pp^{k}
}q^{n-k-1}+p^{k}underbrace{q^{n-k}}_{=qq^{n-k-1}}right) -dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}underbrace{left( pp^{k}q^{n-k-1}
+p^{k}qq^{n-k-1}right) }_{substack{=left( p+qright) p^{k}
q^{n-k-1}=p^{k}q^{n-k-1}\text{(since }p+q=1text{)}}}-dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}}
_{substack{=Q_{n-1,m-1}\text{(by eqref{darij1.pf.l2.0})}}}-dbinom
{n-1}{m-1}p^{m-1}q^{n-m+1}\
& =Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{align*}
This proves Lemma 2. $blacksquare$
Proof of Theorem 1. We have $left( 1-q-pqright) -p^{2}=-left(
p+1right) underbrace{left( p+q-1right) }_{substack{=0\text{(since
}p+q=1text{)}}}=0$, thus $1-q-pq=p^{2}$.
Lemma 2 (applied to $2n+2$ and $n+2$ instead of $n$ and $m$) yields
begin{align}
Q_{2n+2,n+2} & =underbrace{Q_{2n+1,n+1}}_{substack{=Q_{2n,n}-dbinom{2n}
{n}p^{n}q^{left( 2n+1right) -left( n+1right) +1}\text{(by Lemma 2,
applied to }2n+1text{ and }n+1\text{instead of }ntext{ and }mtext{)}
}}-underbrace{dbinom{2n+1}{n+1}}_{substack{=dbinom{2n}{n}+dbinom{2n}
{n+1}\text{(by the recurrence of the binomial coefficients)}}}p^{n+1}
underbrace{q^{left( 2n+2right) -left( n+2right) +1}}_{=q^{n+1}
}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}underbrace{q^{left( 2n+1right) -left(
n+1right) +1}}_{=q^{n+1}}-underbrace{left( dbinom{2n}{n}+dbinom
{2n}{n+1}right) p^{n+1}q^{n+1}}_{=dbinom{2n}{n}p^{n+1}q^{n+1}+dbinom
{2n}{n+1}p^{n+1}q^{n+1}}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-left( dbinom{2n}{n}p^{n+1}
q^{n+1}+dbinom{2n}{n+1}p^{n+1}q^{n+1}right) nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
label{darij1.pf.t1.1}
tag{3}
end{align}
But the definition of $Q_{2n,n+1}$ yields $Q_{2n,n+1}=sum_{k=n+1}^{2n}
dbinom{2n}{k}p^{k}q^{2n-k}$. Meanwhile, the definition of $Q_{2n,n}$ yields
begin{align*}
Q_{2n,n} & =sum_{k=n}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}=underbrace{sum
_{k=n+1}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}}_{=Q_{2n,n+1}}+dbinom{2n}{n}
p^{n}underbrace{q^{2n-n}}_{=q^{n}}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}.
end{align*}
Thus,
eqref{darij1.pf.t1.1} becomes
begin{align*}
Q_{2n+2,n+2} & =underbrace{Q_{2n,n}}_{=Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}
}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}
{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom
{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( p^{n}q^{n}-p^{n}
q^{n+1}-p^{n+1}q^{n+1}right) }_{=left( 1-q-pqright) p^{n}q^{n}}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( 1-q-pqright) }_{=p^{2}}
p^{n}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{p^{2}p^{n}}_{=p^{n+2}}q^{n}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
end{align*}
This proves Theorem 1. $blacksquare$
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
$endgroup$
add a comment |
$begingroup$
Forget the requirement that $0<p<q<1$; it is unnecessary. So let $p$ and $q$
be any reals satisfying $p+q=1$.
Also, let me extend the stage a little bit:
Definition. Let $m$ and $n$ be integers such that $ngeq mgeq0$. Then, we
let
begin{equation}
Q_{n,m}=sum_{k=m}^{n}dbinom{n}{k}p^{k}q^{n-k}.
end{equation}
With this definition, your $P_{2n}$ is $Q_{2n,n+1}$, while your $P_{2n+2}$ is
$Q_{2n+2,n+2}$. Thus, your claim becomes:
Theorem 1. Let $n$ be a nonnegative integer. Then,
begin{equation}
Q_{2n+2,n+2}=Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}
p^{n+1}q^{n+1}.
end{equation}
I shall derive this from the following:
Lemma 2. Let $m$ and $n$ be integers such that $ngeq mgeq1$. Then,
begin{equation}
Q_{n,m}=Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{equation}
Proof of Lemma 2. From $ngeq mgeq1$, we obtain $n-1geq m-1geq0$. Now,
the definition of $Q_{n-1,m-1}$ yields
begin{equation}
Q_{n-1,m-1}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}underbrace{q^{n-1-k}
}_{=q^{n-k-1}}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}.
label{darij1.pf.l2.0}
tag{1}
end{equation}
But the definition of $Q_{n,m}$ yields
begin{align}
Q_{n,m} & =sum_{k=m}^{n}underbrace{dbinom{n}{k}}_{substack{=dbinom
{n-1}{k-1}+dbinom{n-1}{k}\text{(by the recurrence of the binomial
coefficients)}}}p^{k}q^{n-k}=sum_{k=m}^{n}left( dbinom{n-1}{k-1}
+dbinom{n-1}{k}right) p^{k}q^{n-k}nonumber\
& =sum_{k=m}^{n}dbinom{n-1}{k-1}p^{k}q^{n-k}+sum_{k=m}^{n}dbinom{n-1}
{k}p^{k}q^{n-k}nonumber\
& =sum_{k=m-1}^{n-1}underbrace{dbinom{n-1}{left( k+1right) -1}
}_{=dbinom{n-1}{k}}p^{k+1}underbrace{q^{n-left( k+1right) }}
_{=q^{n-k-1}}+sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}nonumber\
& qquadleft( text{here, we have substituted }k+1text{ for }ktext{ in
the first sum}right) nonumber\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum_{k=m}^{n}
dbinom{n-1}{k}p^{k}q^{n-k}.
label{darij1.pf.l2.1}
tag{2}
end{align}
But we have
begin{equation}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m}^{n}dbinom{n-1}{k}
p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }
end{equation}
and
begin{align*}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k} & =sum_{k=m-1}^{n-1}dbinom
{n-1}{k}p^{k}q^{n-k}+underbrace{dbinom{n-1}{n}}_{substack{=0\text{(since
}n-1geq0text{ and }n-1<ntext{)}}}p^{n}q^{n-n}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{align*}
Comparing these two equalities, we obtain
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left(
m-1right) }=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{equation}
Thus,
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }.
end{equation}
Hence, eqref{darij1.pf.l2.1} becomes
begin{align*}
Q_{n,m} & =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+underbrace{sum
_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum
_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}left( p^{k+1}q^{n-k-1}+p^{k}q^{n-k}right) }-dbinom{n-1}{m-1}
p^{m-1}underbrace{q^{n-left( m-1right) }}_{=q^{n-m+1}}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}left( underbrace{p^{k+1}}_{=pp^{k}
}q^{n-k-1}+p^{k}underbrace{q^{n-k}}_{=qq^{n-k-1}}right) -dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}underbrace{left( pp^{k}q^{n-k-1}
+p^{k}qq^{n-k-1}right) }_{substack{=left( p+qright) p^{k}
q^{n-k-1}=p^{k}q^{n-k-1}\text{(since }p+q=1text{)}}}-dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}}
_{substack{=Q_{n-1,m-1}\text{(by eqref{darij1.pf.l2.0})}}}-dbinom
{n-1}{m-1}p^{m-1}q^{n-m+1}\
& =Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{align*}
This proves Lemma 2. $blacksquare$
Proof of Theorem 1. We have $left( 1-q-pqright) -p^{2}=-left(
p+1right) underbrace{left( p+q-1right) }_{substack{=0\text{(since
}p+q=1text{)}}}=0$, thus $1-q-pq=p^{2}$.
Lemma 2 (applied to $2n+2$ and $n+2$ instead of $n$ and $m$) yields
begin{align}
Q_{2n+2,n+2} & =underbrace{Q_{2n+1,n+1}}_{substack{=Q_{2n,n}-dbinom{2n}
{n}p^{n}q^{left( 2n+1right) -left( n+1right) +1}\text{(by Lemma 2,
applied to }2n+1text{ and }n+1\text{instead of }ntext{ and }mtext{)}
}}-underbrace{dbinom{2n+1}{n+1}}_{substack{=dbinom{2n}{n}+dbinom{2n}
{n+1}\text{(by the recurrence of the binomial coefficients)}}}p^{n+1}
underbrace{q^{left( 2n+2right) -left( n+2right) +1}}_{=q^{n+1}
}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}underbrace{q^{left( 2n+1right) -left(
n+1right) +1}}_{=q^{n+1}}-underbrace{left( dbinom{2n}{n}+dbinom
{2n}{n+1}right) p^{n+1}q^{n+1}}_{=dbinom{2n}{n}p^{n+1}q^{n+1}+dbinom
{2n}{n+1}p^{n+1}q^{n+1}}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-left( dbinom{2n}{n}p^{n+1}
q^{n+1}+dbinom{2n}{n+1}p^{n+1}q^{n+1}right) nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
label{darij1.pf.t1.1}
tag{3}
end{align}
But the definition of $Q_{2n,n+1}$ yields $Q_{2n,n+1}=sum_{k=n+1}^{2n}
dbinom{2n}{k}p^{k}q^{2n-k}$. Meanwhile, the definition of $Q_{2n,n}$ yields
begin{align*}
Q_{2n,n} & =sum_{k=n}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}=underbrace{sum
_{k=n+1}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}}_{=Q_{2n,n+1}}+dbinom{2n}{n}
p^{n}underbrace{q^{2n-n}}_{=q^{n}}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}.
end{align*}
Thus,
eqref{darij1.pf.t1.1} becomes
begin{align*}
Q_{2n+2,n+2} & =underbrace{Q_{2n,n}}_{=Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}
}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}
{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom
{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( p^{n}q^{n}-p^{n}
q^{n+1}-p^{n+1}q^{n+1}right) }_{=left( 1-q-pqright) p^{n}q^{n}}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( 1-q-pqright) }_{=p^{2}}
p^{n}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{p^{2}p^{n}}_{=p^{n+2}}q^{n}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
end{align*}
This proves Theorem 1. $blacksquare$
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
$endgroup$
Forget the requirement that $0<p<q<1$; it is unnecessary. So let $p$ and $q$
be any reals satisfying $p+q=1$.
Also, let me extend the stage a little bit:
Definition. Let $m$ and $n$ be integers such that $ngeq mgeq0$. Then, we
let
begin{equation}
Q_{n,m}=sum_{k=m}^{n}dbinom{n}{k}p^{k}q^{n-k}.
end{equation}
With this definition, your $P_{2n}$ is $Q_{2n,n+1}$, while your $P_{2n+2}$ is
$Q_{2n+2,n+2}$. Thus, your claim becomes:
Theorem 1. Let $n$ be a nonnegative integer. Then,
begin{equation}
Q_{2n+2,n+2}=Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}
p^{n+1}q^{n+1}.
end{equation}
I shall derive this from the following:
Lemma 2. Let $m$ and $n$ be integers such that $ngeq mgeq1$. Then,
begin{equation}
Q_{n,m}=Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{equation}
Proof of Lemma 2. From $ngeq mgeq1$, we obtain $n-1geq m-1geq0$. Now,
the definition of $Q_{n-1,m-1}$ yields
begin{equation}
Q_{n-1,m-1}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}underbrace{q^{n-1-k}
}_{=q^{n-k-1}}=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}.
label{darij1.pf.l2.0}
tag{1}
end{equation}
But the definition of $Q_{n,m}$ yields
begin{align}
Q_{n,m} & =sum_{k=m}^{n}underbrace{dbinom{n}{k}}_{substack{=dbinom
{n-1}{k-1}+dbinom{n-1}{k}\text{(by the recurrence of the binomial
coefficients)}}}p^{k}q^{n-k}=sum_{k=m}^{n}left( dbinom{n-1}{k-1}
+dbinom{n-1}{k}right) p^{k}q^{n-k}nonumber\
& =sum_{k=m}^{n}dbinom{n-1}{k-1}p^{k}q^{n-k}+sum_{k=m}^{n}dbinom{n-1}
{k}p^{k}q^{n-k}nonumber\
& =sum_{k=m-1}^{n-1}underbrace{dbinom{n-1}{left( k+1right) -1}
}_{=dbinom{n-1}{k}}p^{k+1}underbrace{q^{n-left( k+1right) }}
_{=q^{n-k-1}}+sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}nonumber\
& qquadleft( text{here, we have substituted }k+1text{ for }ktext{ in
the first sum}right) nonumber\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum_{k=m}^{n}
dbinom{n-1}{k}p^{k}q^{n-k}.
label{darij1.pf.l2.1}
tag{2}
end{align}
But we have
begin{equation}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m}^{n}dbinom{n-1}{k}
p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }
end{equation}
and
begin{align*}
sum_{k=m-1}^{n}dbinom{n-1}{k}p^{k}q^{n-k} & =sum_{k=m-1}^{n-1}dbinom
{n-1}{k}p^{k}q^{n-k}+underbrace{dbinom{n-1}{n}}_{substack{=0\text{(since
}n-1geq0text{ and }n-1<ntext{)}}}p^{n}q^{n-n}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{align*}
Comparing these two equalities, we obtain
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}+dbinom{n-1}{m-1}p^{m-1}q^{n-left(
m-1right) }=sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}.
end{equation}
Thus,
begin{equation}
sum_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }.
end{equation}
Hence, eqref{darij1.pf.l2.1} becomes
begin{align*}
Q_{n,m} & =sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+underbrace{sum
_{k=m}^{n}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}p^{k}q^{n-k}-dbinom{n-1}{m-1}p^{m-1}q^{n-left( m-1right) }}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k+1}q^{n-k-1}+sum
_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k}}_{=sum_{k=m-1}^{n-1}dbinom{n-1}
{k}left( p^{k+1}q^{n-k-1}+p^{k}q^{n-k}right) }-dbinom{n-1}{m-1}
p^{m-1}underbrace{q^{n-left( m-1right) }}_{=q^{n-m+1}}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}left( underbrace{p^{k+1}}_{=pp^{k}
}q^{n-k-1}+p^{k}underbrace{q^{n-k}}_{=qq^{n-k-1}}right) -dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =sum_{k=m-1}^{n-1}dbinom{n-1}{k}underbrace{left( pp^{k}q^{n-k-1}
+p^{k}qq^{n-k-1}right) }_{substack{=left( p+qright) p^{k}
q^{n-k-1}=p^{k}q^{n-k-1}\text{(since }p+q=1text{)}}}-dbinom{n-1}
{m-1}p^{m-1}q^{n-m+1}\
& =underbrace{sum_{k=m-1}^{n-1}dbinom{n-1}{k}p^{k}q^{n-k-1}}
_{substack{=Q_{n-1,m-1}\text{(by eqref{darij1.pf.l2.0})}}}-dbinom
{n-1}{m-1}p^{m-1}q^{n-m+1}\
& =Q_{n-1,m-1}-dbinom{n-1}{m-1}p^{m-1}q^{n-m+1}.
end{align*}
This proves Lemma 2. $blacksquare$
Proof of Theorem 1. We have $left( 1-q-pqright) -p^{2}=-left(
p+1right) underbrace{left( p+q-1right) }_{substack{=0\text{(since
}p+q=1text{)}}}=0$, thus $1-q-pq=p^{2}$.
Lemma 2 (applied to $2n+2$ and $n+2$ instead of $n$ and $m$) yields
begin{align}
Q_{2n+2,n+2} & =underbrace{Q_{2n+1,n+1}}_{substack{=Q_{2n,n}-dbinom{2n}
{n}p^{n}q^{left( 2n+1right) -left( n+1right) +1}\text{(by Lemma 2,
applied to }2n+1text{ and }n+1\text{instead of }ntext{ and }mtext{)}
}}-underbrace{dbinom{2n+1}{n+1}}_{substack{=dbinom{2n}{n}+dbinom{2n}
{n+1}\text{(by the recurrence of the binomial coefficients)}}}p^{n+1}
underbrace{q^{left( 2n+2right) -left( n+2right) +1}}_{=q^{n+1}
}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}underbrace{q^{left( 2n+1right) -left(
n+1right) +1}}_{=q^{n+1}}-underbrace{left( dbinom{2n}{n}+dbinom
{2n}{n+1}right) p^{n+1}q^{n+1}}_{=dbinom{2n}{n}p^{n+1}q^{n+1}+dbinom
{2n}{n+1}p^{n+1}q^{n+1}}nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-left( dbinom{2n}{n}p^{n+1}
q^{n+1}+dbinom{2n}{n+1}p^{n+1}q^{n+1}right) nonumber\
& =Q_{2n,n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
label{darij1.pf.t1.1}
tag{3}
end{align}
But the definition of $Q_{2n,n+1}$ yields $Q_{2n,n+1}=sum_{k=n+1}^{2n}
dbinom{2n}{k}p^{k}q^{2n-k}$. Meanwhile, the definition of $Q_{2n,n}$ yields
begin{align*}
Q_{2n,n} & =sum_{k=n}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}=underbrace{sum
_{k=n+1}^{2n}dbinom{2n}{k}p^{k}q^{2n-k}}_{=Q_{2n,n+1}}+dbinom{2n}{n}
p^{n}underbrace{q^{2n-n}}_{=q^{n}}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}.
end{align*}
Thus,
eqref{darij1.pf.t1.1} becomes
begin{align*}
Q_{2n+2,n+2} & =underbrace{Q_{2n,n}}_{=Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}
}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}
{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n}q^{n}-dbinom{2n}{n}p^{n}q^{n+1}-dbinom
{2n}{n}p^{n+1}q^{n+1}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( p^{n}q^{n}-p^{n}
q^{n+1}-p^{n+1}q^{n+1}right) }_{=left( 1-q-pqright) p^{n}q^{n}}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{left( 1-q-pqright) }_{=p^{2}}
p^{n}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}underbrace{p^{2}p^{n}}_{=p^{n+2}}q^{n}
-dbinom{2n}{n+1}p^{n+1}q^{n+1}\
& =Q_{2n,n+1}+dbinom{2n}{n}p^{n+2}q^{n}-dbinom{2n}{n+1}p^{n+1}q^{n+1}.
end{align*}
This proves Theorem 1. $blacksquare$
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
answered Jan 13 at 13:12
darij grinbergdarij grinberg
10.5k33062
10.5k33062
add a comment |
add a comment |
$begingroup$
We have for the first sum
$$S = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = sum_{k=0}^{n-1} {2nchoose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
sum_{k=0}^n {2n+2choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
sum_{k=0}^n {2n+2choose n-k} p^{k} q^{-k}
\ = p^{n+2} q^n
sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{kge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} frac{1}{1-pz/q}
\ = p^2 [z^n] (1+pqz)^{2n+2} frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] z (1+pqz)^{2n} frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\ = p^2 p^n q^n {2nchoose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\ = p^{n+2} q^n {2nchoose n}
- p^2 q^2 p^{n-1} q^{n-1} {2nchoose n-1}.$$
This is indeed
$$bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2nchoose n}
- p^{n+1} q^{n+1} {2nchoose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
add a comment |
$begingroup$
We have for the first sum
$$S = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = sum_{k=0}^{n-1} {2nchoose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
sum_{k=0}^n {2n+2choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
sum_{k=0}^n {2n+2choose n-k} p^{k} q^{-k}
\ = p^{n+2} q^n
sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{kge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} frac{1}{1-pz/q}
\ = p^2 [z^n] (1+pqz)^{2n+2} frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] z (1+pqz)^{2n} frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\ = p^2 p^n q^n {2nchoose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\ = p^{n+2} q^n {2nchoose n}
- p^2 q^2 p^{n-1} q^{n-1} {2nchoose n-1}.$$
This is indeed
$$bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2nchoose n}
- p^{n+1} q^{n+1} {2nchoose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
add a comment |
$begingroup$
We have for the first sum
$$S = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = sum_{k=0}^{n-1} {2nchoose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
sum_{k=0}^n {2n+2choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
sum_{k=0}^n {2n+2choose n-k} p^{k} q^{-k}
\ = p^{n+2} q^n
sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{kge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} frac{1}{1-pz/q}
\ = p^2 [z^n] (1+pqz)^{2n+2} frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] z (1+pqz)^{2n} frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\ = p^2 p^n q^n {2nchoose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\ = p^{n+2} q^n {2nchoose n}
- p^2 q^2 p^{n-1} q^{n-1} {2nchoose n-1}.$$
This is indeed
$$bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2nchoose n}
- p^{n+1} q^{n+1} {2nchoose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
We have for the first sum
$$S = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = sum_{k=0}^{n-1} {2nchoose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = sum_{k=0}^n {2n+2choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2nchoose n} p^{n+2} q^n - {2nchoose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
sum_{k=0}^n {2n+2choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
sum_{k=0}^n {2n+2choose n-k} p^{k} q^{-k}
\ = p^{n+2} q^n
sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
sum_{kge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} frac{1}{1-pz/q}
\ = p^2 [z^n] (1+pqz)^{2n+2} frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] z (1+pqz)^{2n} frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} frac{1}{1-p^2z}
\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\ = p^2 p^n q^n {2nchoose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\ = p^{n+2} q^n {2nchoose n}
- p^2 q^2 p^{n-1} q^{n-1} {2nchoose n-1}.$$
This is indeed
$$bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2nchoose n}
- p^{n+1} q^{n+1} {2nchoose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
edited Jan 13 at 16:11
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
answered Jan 13 at 15:55
Marko RiedelMarko Riedel
39.7k339108
39.7k339108
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$begingroup$
This is perhaps true. It reduces to $binom{2 n}{n+1} left((p-1)^2-, _2F_1left(1,1-n;n+2;frac{p}{p-1}right)right)+(p-1) p binom{2 n}{n}=(p-1) p binom{2 (n+1)}{n+2} , _2F_1left(1,-n;n+3;frac{p}{p-1}right)$. Though I do not know how to prove this.
$endgroup$
– ablmf
Jan 13 at 10:30