Proof $sum_{k=0}^infty binom{k}{n-k} = f_{n+1}$ where $f_n$ is the n-th Fibonacci-number












5












$begingroup$


In our combinatorics script it is written, that



$$sum_{k=0}^infty binom{k}{n-k} = f_{n+1}$$ where $f_n$ is the n-th Fibonacci-number.



Apparently this can be proven through the generating function



$$A(x) = sum_{n=0}^infty x^n sum_{k=0}^infty binom{k}{n-k}$$



I've looked on stackexchange math and on the internet, but I couldn't find a proof.



I know that $$sum_{kge0} binom{n-k}k=f_{n+1}$$



and it can be proven through this



$$sum_{kge0} binom{n+1-k}k=sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-k}{k-1}=
sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-1-(k-1)}{k-1}=
sum_{kge0} binom{n-k}k + sum_{jge0} binom{n-1-j}{j}= f_{n+1}+f_n = f_{n+2}.$$



But I don't know how its done for the first case.






combinatorics proof-writing binomial-coefficients fibonacci-numbers







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    In our combinatorics script it is written, that



    $$sum_{k=0}^infty binom{k}{n-k} = f_{n+1}$$ where $f_n$ is the n-th Fibonacci-number.



    Apparently this can be proven through the generating function



    $$A(x) = sum_{n=0}^infty x^n sum_{k=0}^infty binom{k}{n-k}$$



    I've looked on stackexchange math and on the internet, but I couldn't find a proof.



    I know that $$sum_{kge0} binom{n-k}k=f_{n+1}$$



    and it can be proven through this



    $$sum_{kge0} binom{n+1-k}k=sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-k}{k-1}=
    sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-1-(k-1)}{k-1}=
    sum_{kge0} binom{n-k}k + sum_{jge0} binom{n-1-j}{j}= f_{n+1}+f_n = f_{n+2}.$$



    But I don't know how its done for the first case.






    combinatorics proof-writing binomial-coefficients fibonacci-numbers







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      In our combinatorics script it is written, that



      $$sum_{k=0}^infty binom{k}{n-k} = f_{n+1}$$ where $f_n$ is the n-th Fibonacci-number.



      Apparently this can be proven through the generating function



      $$A(x) = sum_{n=0}^infty x^n sum_{k=0}^infty binom{k}{n-k}$$



      I've looked on stackexchange math and on the internet, but I couldn't find a proof.



      I know that $$sum_{kge0} binom{n-k}k=f_{n+1}$$



      and it can be proven through this



      $$sum_{kge0} binom{n+1-k}k=sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-k}{k-1}=
      sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-1-(k-1)}{k-1}=
      sum_{kge0} binom{n-k}k + sum_{jge0} binom{n-1-j}{j}= f_{n+1}+f_n = f_{n+2}.$$



      But I don't know how its done for the first case.






      combinatorics proof-writing binomial-coefficients fibonacci-numbers







      share|cite|improve this question









      $endgroup$




      In our combinatorics script it is written, that



      $$sum_{k=0}^infty binom{k}{n-k} = f_{n+1}$$ where $f_n$ is the n-th Fibonacci-number.



      Apparently this can be proven through the generating function



      $$A(x) = sum_{n=0}^infty x^n sum_{k=0}^infty binom{k}{n-k}$$



      I've looked on stackexchange math and on the internet, but I couldn't find a proof.



      I know that $$sum_{kge0} binom{n-k}k=f_{n+1}$$



      and it can be proven through this



      $$sum_{kge0} binom{n+1-k}k=sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-k}{k-1}=
      sum_{kge0} binom{n-k}k + sum_{kge0} binom{n-1-(k-1)}{k-1}=
      sum_{kge0} binom{n-k}k + sum_{jge0} binom{n-1-j}{j}= f_{n+1}+f_n = f_{n+2}.$$



      But I don't know how its done for the first case.






      combinatorics proof-writing binomial-coefficients fibonacci-numbers




      combinatorics proof-writing binomial-coefficients fibonacci-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 21:13









      NotEinsteinNotEinstein

      2146




      2146






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          begin{align}
          A(x)
          &= sum_{nge 0} x^nsum_{kge 0} binom{k}{n-k}
          \&= sum_{kge 0} sum_{nge 0} binom{k}{n-k}x^n
          \&= sum_{kge 0} x^ksum_{nge k} binom{k}{n-k}x^{n-k}
          \&= sum_{kge 0} x^ksum_{nge 0} binom{k}{n}x^{n}
          \&= sum_{kge 0} x^k(1+x)^k
          \&= frac1{1-(x+x^2)}
          end{align}
          All that remains is to recall $sum_{nge 0} f_nx^n=frac{x}{1-x-x^2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much. Can you tell me how you get to (1+x)^k?
            $endgroup$
            – NotEinstein
            Jan 24 at 21:52








          • 1




            $begingroup$
            @NotEinstein Use the binomial theorem. See edit for more detail.
            $endgroup$
            – Mike Earnest
            Jan 24 at 21:54



















          1












          $begingroup$

          Both $binom{k}{n - k}$ and $binom{n - k}{k}$ are only non-zero if $0 le k le n$ (so the sums are finite). The sum $sum_{k} binom{n-k}k$ can be written as



          $$ binom{n}{0} + binom{n - 1}{1} + binom{n - 2}{2} + cdots + binom{n - n}{n}. $$



          If you reverse the order (which corresponds to the involution $k leftrightarrow n - k$), you get



          $$ binom{0}{n} + binom{1}{n - 1} + binom{2}{n-2} + cdots + binom{n}{n-n},$$



          which is the sum $sum_k binom{k}{n-k}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
            $endgroup$
            – darij grinberg
            Jan 24 at 21:57










          • $begingroup$
            @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:08










          • $begingroup$
            The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
            $endgroup$
            – darij grinberg
            Jan 24 at 23:30










          • $begingroup$
            @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:40










          • $begingroup$
            @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
            $endgroup$
            – Mike Earnest
            Feb 15 at 17:23











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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          begin{align}
          A(x)
          &= sum_{nge 0} x^nsum_{kge 0} binom{k}{n-k}
          \&= sum_{kge 0} sum_{nge 0} binom{k}{n-k}x^n
          \&= sum_{kge 0} x^ksum_{nge k} binom{k}{n-k}x^{n-k}
          \&= sum_{kge 0} x^ksum_{nge 0} binom{k}{n}x^{n}
          \&= sum_{kge 0} x^k(1+x)^k
          \&= frac1{1-(x+x^2)}
          end{align}
          All that remains is to recall $sum_{nge 0} f_nx^n=frac{x}{1-x-x^2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much. Can you tell me how you get to (1+x)^k?
            $endgroup$
            – NotEinstein
            Jan 24 at 21:52








          • 1




            $begingroup$
            @NotEinstein Use the binomial theorem. See edit for more detail.
            $endgroup$
            – Mike Earnest
            Jan 24 at 21:54
















          7












          $begingroup$

          begin{align}
          A(x)
          &= sum_{nge 0} x^nsum_{kge 0} binom{k}{n-k}
          \&= sum_{kge 0} sum_{nge 0} binom{k}{n-k}x^n
          \&= sum_{kge 0} x^ksum_{nge k} binom{k}{n-k}x^{n-k}
          \&= sum_{kge 0} x^ksum_{nge 0} binom{k}{n}x^{n}
          \&= sum_{kge 0} x^k(1+x)^k
          \&= frac1{1-(x+x^2)}
          end{align}
          All that remains is to recall $sum_{nge 0} f_nx^n=frac{x}{1-x-x^2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much. Can you tell me how you get to (1+x)^k?
            $endgroup$
            – NotEinstein
            Jan 24 at 21:52








          • 1




            $begingroup$
            @NotEinstein Use the binomial theorem. See edit for more detail.
            $endgroup$
            – Mike Earnest
            Jan 24 at 21:54














          7












          7








          7





          $begingroup$

          begin{align}
          A(x)
          &= sum_{nge 0} x^nsum_{kge 0} binom{k}{n-k}
          \&= sum_{kge 0} sum_{nge 0} binom{k}{n-k}x^n
          \&= sum_{kge 0} x^ksum_{nge k} binom{k}{n-k}x^{n-k}
          \&= sum_{kge 0} x^ksum_{nge 0} binom{k}{n}x^{n}
          \&= sum_{kge 0} x^k(1+x)^k
          \&= frac1{1-(x+x^2)}
          end{align}
          All that remains is to recall $sum_{nge 0} f_nx^n=frac{x}{1-x-x^2}$.






          share|cite|improve this answer











          $endgroup$



          begin{align}
          A(x)
          &= sum_{nge 0} x^nsum_{kge 0} binom{k}{n-k}
          \&= sum_{kge 0} sum_{nge 0} binom{k}{n-k}x^n
          \&= sum_{kge 0} x^ksum_{nge k} binom{k}{n-k}x^{n-k}
          \&= sum_{kge 0} x^ksum_{nge 0} binom{k}{n}x^{n}
          \&= sum_{kge 0} x^k(1+x)^k
          \&= frac1{1-(x+x^2)}
          end{align}
          All that remains is to recall $sum_{nge 0} f_nx^n=frac{x}{1-x-x^2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 24 at 21:54

























          answered Jan 24 at 21:31









          Mike EarnestMike Earnest

          24.3k22151




          24.3k22151












          • $begingroup$
            Thank you very much. Can you tell me how you get to (1+x)^k?
            $endgroup$
            – NotEinstein
            Jan 24 at 21:52








          • 1




            $begingroup$
            @NotEinstein Use the binomial theorem. See edit for more detail.
            $endgroup$
            – Mike Earnest
            Jan 24 at 21:54


















          • $begingroup$
            Thank you very much. Can you tell me how you get to (1+x)^k?
            $endgroup$
            – NotEinstein
            Jan 24 at 21:52








          • 1




            $begingroup$
            @NotEinstein Use the binomial theorem. See edit for more detail.
            $endgroup$
            – Mike Earnest
            Jan 24 at 21:54
















          $begingroup$
          Thank you very much. Can you tell me how you get to (1+x)^k?
          $endgroup$
          – NotEinstein
          Jan 24 at 21:52






          $begingroup$
          Thank you very much. Can you tell me how you get to (1+x)^k?
          $endgroup$
          – NotEinstein
          Jan 24 at 21:52






          1




          1




          $begingroup$
          @NotEinstein Use the binomial theorem. See edit for more detail.
          $endgroup$
          – Mike Earnest
          Jan 24 at 21:54




          $begingroup$
          @NotEinstein Use the binomial theorem. See edit for more detail.
          $endgroup$
          – Mike Earnest
          Jan 24 at 21:54











          1












          $begingroup$

          Both $binom{k}{n - k}$ and $binom{n - k}{k}$ are only non-zero if $0 le k le n$ (so the sums are finite). The sum $sum_{k} binom{n-k}k$ can be written as



          $$ binom{n}{0} + binom{n - 1}{1} + binom{n - 2}{2} + cdots + binom{n - n}{n}. $$



          If you reverse the order (which corresponds to the involution $k leftrightarrow n - k$), you get



          $$ binom{0}{n} + binom{1}{n - 1} + binom{2}{n-2} + cdots + binom{n}{n-n},$$



          which is the sum $sum_k binom{k}{n-k}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
            $endgroup$
            – darij grinberg
            Jan 24 at 21:57










          • $begingroup$
            @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:08










          • $begingroup$
            The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
            $endgroup$
            – darij grinberg
            Jan 24 at 23:30










          • $begingroup$
            @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:40










          • $begingroup$
            @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
            $endgroup$
            – Mike Earnest
            Feb 15 at 17:23
















          1












          $begingroup$

          Both $binom{k}{n - k}$ and $binom{n - k}{k}$ are only non-zero if $0 le k le n$ (so the sums are finite). The sum $sum_{k} binom{n-k}k$ can be written as



          $$ binom{n}{0} + binom{n - 1}{1} + binom{n - 2}{2} + cdots + binom{n - n}{n}. $$



          If you reverse the order (which corresponds to the involution $k leftrightarrow n - k$), you get



          $$ binom{0}{n} + binom{1}{n - 1} + binom{2}{n-2} + cdots + binom{n}{n-n},$$



          which is the sum $sum_k binom{k}{n-k}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
            $endgroup$
            – darij grinberg
            Jan 24 at 21:57










          • $begingroup$
            @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:08










          • $begingroup$
            The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
            $endgroup$
            – darij grinberg
            Jan 24 at 23:30










          • $begingroup$
            @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:40










          • $begingroup$
            @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
            $endgroup$
            – Mike Earnest
            Feb 15 at 17:23














          1












          1








          1





          $begingroup$

          Both $binom{k}{n - k}$ and $binom{n - k}{k}$ are only non-zero if $0 le k le n$ (so the sums are finite). The sum $sum_{k} binom{n-k}k$ can be written as



          $$ binom{n}{0} + binom{n - 1}{1} + binom{n - 2}{2} + cdots + binom{n - n}{n}. $$



          If you reverse the order (which corresponds to the involution $k leftrightarrow n - k$), you get



          $$ binom{0}{n} + binom{1}{n - 1} + binom{2}{n-2} + cdots + binom{n}{n-n},$$



          which is the sum $sum_k binom{k}{n-k}$.






          share|cite|improve this answer









          $endgroup$



          Both $binom{k}{n - k}$ and $binom{n - k}{k}$ are only non-zero if $0 le k le n$ (so the sums are finite). The sum $sum_{k} binom{n-k}k$ can be written as



          $$ binom{n}{0} + binom{n - 1}{1} + binom{n - 2}{2} + cdots + binom{n - n}{n}. $$



          If you reverse the order (which corresponds to the involution $k leftrightarrow n - k$), you get



          $$ binom{0}{n} + binom{1}{n - 1} + binom{2}{n-2} + cdots + binom{n}{n-n},$$



          which is the sum $sum_k binom{k}{n-k}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 21:23









          Trevor GunnTrevor Gunn

          14.8k32047




          14.8k32047












          • $begingroup$
            $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
            $endgroup$
            – darij grinberg
            Jan 24 at 21:57










          • $begingroup$
            @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:08










          • $begingroup$
            The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
            $endgroup$
            – darij grinberg
            Jan 24 at 23:30










          • $begingroup$
            @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:40










          • $begingroup$
            @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
            $endgroup$
            – Mike Earnest
            Feb 15 at 17:23


















          • $begingroup$
            $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
            $endgroup$
            – darij grinberg
            Jan 24 at 21:57










          • $begingroup$
            @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:08










          • $begingroup$
            The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
            $endgroup$
            – darij grinberg
            Jan 24 at 23:30










          • $begingroup$
            @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
            $endgroup$
            – Trevor Gunn
            Jan 24 at 23:40










          • $begingroup$
            @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
            $endgroup$
            – Mike Earnest
            Feb 15 at 17:23
















          $begingroup$
          $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
          $endgroup$
          – darij grinberg
          Jan 24 at 21:57




          $begingroup$
          $dbinom{n-k}{k}$ can be nonzero when $k$ is larger than $n$. This is a major annoyance and every combinatorialist has stumbled at least one across it.
          $endgroup$
          – darij grinberg
          Jan 24 at 21:57












          $begingroup$
          @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
          $endgroup$
          – Trevor Gunn
          Jan 24 at 23:08




          $begingroup$
          @Darij I'm not seeing it. Are you possibly thinking of a different but similar looking quantity? Combinatorially, $F_n$ is the number of ways to write $n$ as a sum of ones and twos and $binom{n - k}{k}$ is the number of ways with $n - 2k$ ones and $k$ twos. Certainly you can't have more than $n/2$ twos.
          $endgroup$
          – Trevor Gunn
          Jan 24 at 23:08












          $begingroup$
          The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
          $endgroup$
          – darij grinberg
          Jan 24 at 23:30




          $begingroup$
          The combinatorial interpretation requires the "numerator" to be nonnegative, while $dbinom{x}{k} := dfrac{xleft(x-1right)cdotsleft(x-k+1right)}{k!}$ gives nonzero values for negative $x$.
          $endgroup$
          – darij grinberg
          Jan 24 at 23:30












          $begingroup$
          @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
          $endgroup$
          – Trevor Gunn
          Jan 24 at 23:40




          $begingroup$
          @Darij I see that, but the sum in the question doesn't make sense if you allow those values. In this context, it is necessary to require $binom{n}{k} = 0$ unless $n ge 0$ and $0 le k le n$.
          $endgroup$
          – Trevor Gunn
          Jan 24 at 23:40












          $begingroup$
          @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
          $endgroup$
          – Mike Earnest
          Feb 15 at 17:23




          $begingroup$
          @darijgrinberg There are two equally valid generalizations of $binom{n}k$ to real $n,k$. One is $$binom{n}k=begin{cases}(n)_k/k! & kin mathbb N\0&text{otherwise}end{cases}$$ This is the generalization where $binom{n}k$ can be zero when $kge 0$ but $n$ is negative. The other generalization is $$binom{n}k=begin{cases}(n)_{n-k}/(n-k)! & n-kin mathbb N\0&text{otherwise}end{cases}$$ For this generalization, you do have that $sum_{k}binom{n-k}{k}$ is a finite sum equal to $F_{n+1}$. This works even for negative $n$. See en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
          $endgroup$
          – Mike Earnest
          Feb 15 at 17:23


















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