How to prove combinatorial $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$












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I have problems solving the following formula for even positive integers $n$:



$$sum_{k=0}^{n/2} {nchoose 2k} = sum_{k=0}^{n/2 - 1} {nchoose 2k+1} = 2^{n-1}$$



I tried to prove it by induction but it didn't work. Is there any combinatorial proof?






combinatorics summation binomial-coefficients







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  • 1




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    You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
    $endgroup$
    – Severin Schraven
    Jan 24 at 11:15
















3












$begingroup$


I have problems solving the following formula for even positive integers $n$:



$$sum_{k=0}^{n/2} {nchoose 2k} = sum_{k=0}^{n/2 - 1} {nchoose 2k+1} = 2^{n-1}$$



I tried to prove it by induction but it didn't work. Is there any combinatorial proof?






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
    $endgroup$
    – Severin Schraven
    Jan 24 at 11:15














3












3








3


1



$begingroup$


I have problems solving the following formula for even positive integers $n$:



$$sum_{k=0}^{n/2} {nchoose 2k} = sum_{k=0}^{n/2 - 1} {nchoose 2k+1} = 2^{n-1}$$



I tried to prove it by induction but it didn't work. Is there any combinatorial proof?






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$




I have problems solving the following formula for even positive integers $n$:



$$sum_{k=0}^{n/2} {nchoose 2k} = sum_{k=0}^{n/2 - 1} {nchoose 2k+1} = 2^{n-1}$$



I tried to prove it by induction but it didn't work. Is there any combinatorial proof?






combinatorics summation binomial-coefficients




combinatorics summation binomial-coefficients






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share|cite|improve this question













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edited Jan 28 at 2:26









darij grinberg

11.1k33167




11.1k33167










asked Jan 24 at 11:04









NickNick

191




191








  • 1




    $begingroup$
    You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
    $endgroup$
    – Severin Schraven
    Jan 24 at 11:15














  • 1




    $begingroup$
    You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
    $endgroup$
    – Severin Schraven
    Jan 24 at 11:15








1




1




$begingroup$
You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
$endgroup$
– Severin Schraven
Jan 24 at 11:15




$begingroup$
You can use the symmetry of the binomial coefficients to show the first equality. Then you can use the binomial theorem to show that the sum of the first two is equal to $2^n$ and conclude the second equality using the first one.
$endgroup$
– Severin Schraven
Jan 24 at 11:15










3 Answers
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oldest

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3












$begingroup$

Let $S$ denote a set of size $nge 1$. Your first sum is the number of even-size subsets of $S$; the second sum counts the odd-size subsets instead. Fix some $ain S$, then pair each $xsubseteq Sbackslash{a}$ with $xcup{a}$. Since each pair is one odd-sized set and one even-sized set, there are equally many of each, and that's $2^{n-1}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint:
    You can get the result by expanding these two expressions using the binomial theorem
    $$(1-1)^n=0 text{ and } (1+1)^n=2^n.$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      For brevity we prove $$sum_{k=0}^{n} binom{2n}{2k} = sum_{k=0}^{n-1} binom{2n}{2k+1} = 2^{2n-1},hspace{2cm}(*)$$ for $ninmathbb{N}$. It is sufficient to show that $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ since we have
      $$sum_{k=0}^{n} binom{2n}{2k} +sum_{k=0}^{n-1} binom{2n}{2k+1}=sum_{k=0}^{2n} binom{2n}{k} = 2^{2n}.hspace{2cm}(**)$$ Let apply induction for $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$.
      For $n=1$ equality is true. For $n=m$ assume equality is true, namely $sum_{k=0}^{m} binom{2m}{2k}=2^{2m-1}$(Here note that $sum_{k=0}^{m-1} binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
      begin{eqnarray*}sum_{k=0}^{m+1} binom{2m+2}{2k}&=&sum_{k=0}^{m+1}left{ binom{2m}{2k-1}+binom{2m}{2k-2}+binom{2m}{2k}+binom{2m}{2k-1}right}\
      &=&sum_{k=0}^{m+1} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
      &=&sum_{k=0}^{m} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
      &=&sum_{k=0}^{m} binom{2m}{2k}+2sum_{k=0}^{m-1} binom{2m}{2k+1}+sum_{k=0}^{m} binom{2m}{2k}\
      &=&4.2^{2m-1}=2^{2m+1}end{eqnarray*}       by use of $binom{n}{k}=binom{n-1}{k-1}+binom{n-1}{k}$. Hence equality $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ is valid for $ninmathbb{N}$ which implies $(*)$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
        $endgroup$
        – Nick
        Jan 24 at 12:58










      • $begingroup$
        Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
        $endgroup$
        – guest
        Jan 24 at 13:41













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Let $S$ denote a set of size $nge 1$. Your first sum is the number of even-size subsets of $S$; the second sum counts the odd-size subsets instead. Fix some $ain S$, then pair each $xsubseteq Sbackslash{a}$ with $xcup{a}$. Since each pair is one odd-sized set and one even-sized set, there are equally many of each, and that's $2^{n-1}$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Let $S$ denote a set of size $nge 1$. Your first sum is the number of even-size subsets of $S$; the second sum counts the odd-size subsets instead. Fix some $ain S$, then pair each $xsubseteq Sbackslash{a}$ with $xcup{a}$. Since each pair is one odd-sized set and one even-sized set, there are equally many of each, and that's $2^{n-1}$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Let $S$ denote a set of size $nge 1$. Your first sum is the number of even-size subsets of $S$; the second sum counts the odd-size subsets instead. Fix some $ain S$, then pair each $xsubseteq Sbackslash{a}$ with $xcup{a}$. Since each pair is one odd-sized set and one even-sized set, there are equally many of each, and that's $2^{n-1}$.






          share|cite|improve this answer









          $endgroup$



          Let $S$ denote a set of size $nge 1$. Your first sum is the number of even-size subsets of $S$; the second sum counts the odd-size subsets instead. Fix some $ain S$, then pair each $xsubseteq Sbackslash{a}$ with $xcup{a}$. Since each pair is one odd-sized set and one even-sized set, there are equally many of each, and that's $2^{n-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 12:26









          J.G.J.G.

          28.9k22845




          28.9k22845























              2












              $begingroup$

              Hint:
              You can get the result by expanding these two expressions using the binomial theorem
              $$(1-1)^n=0 text{ and } (1+1)^n=2^n.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint:
                You can get the result by expanding these two expressions using the binomial theorem
                $$(1-1)^n=0 text{ and } (1+1)^n=2^n.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint:
                  You can get the result by expanding these two expressions using the binomial theorem
                  $$(1-1)^n=0 text{ and } (1+1)^n=2^n.$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:
                  You can get the result by expanding these two expressions using the binomial theorem
                  $$(1-1)^n=0 text{ and } (1+1)^n=2^n.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 24 at 12:31









                  Teddy38Teddy38

                  2,1912520




                  2,1912520























                      2












                      $begingroup$

                      For brevity we prove $$sum_{k=0}^{n} binom{2n}{2k} = sum_{k=0}^{n-1} binom{2n}{2k+1} = 2^{2n-1},hspace{2cm}(*)$$ for $ninmathbb{N}$. It is sufficient to show that $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ since we have
                      $$sum_{k=0}^{n} binom{2n}{2k} +sum_{k=0}^{n-1} binom{2n}{2k+1}=sum_{k=0}^{2n} binom{2n}{k} = 2^{2n}.hspace{2cm}(**)$$ Let apply induction for $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$.
                      For $n=1$ equality is true. For $n=m$ assume equality is true, namely $sum_{k=0}^{m} binom{2m}{2k}=2^{2m-1}$(Here note that $sum_{k=0}^{m-1} binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
                      begin{eqnarray*}sum_{k=0}^{m+1} binom{2m+2}{2k}&=&sum_{k=0}^{m+1}left{ binom{2m}{2k-1}+binom{2m}{2k-2}+binom{2m}{2k}+binom{2m}{2k-1}right}\
                      &=&sum_{k=0}^{m+1} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+2sum_{k=0}^{m-1} binom{2m}{2k+1}+sum_{k=0}^{m} binom{2m}{2k}\
                      &=&4.2^{2m-1}=2^{2m+1}end{eqnarray*}                       by use of $binom{n}{k}=binom{n-1}{k-1}+binom{n-1}{k}$. Hence equality $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ is valid for $ninmathbb{N}$ which implies $(*)$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                        $endgroup$
                        – Nick
                        Jan 24 at 12:58










                      • $begingroup$
                        Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                        $endgroup$
                        – guest
                        Jan 24 at 13:41


















                      2












                      $begingroup$

                      For brevity we prove $$sum_{k=0}^{n} binom{2n}{2k} = sum_{k=0}^{n-1} binom{2n}{2k+1} = 2^{2n-1},hspace{2cm}(*)$$ for $ninmathbb{N}$. It is sufficient to show that $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ since we have
                      $$sum_{k=0}^{n} binom{2n}{2k} +sum_{k=0}^{n-1} binom{2n}{2k+1}=sum_{k=0}^{2n} binom{2n}{k} = 2^{2n}.hspace{2cm}(**)$$ Let apply induction for $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$.
                      For $n=1$ equality is true. For $n=m$ assume equality is true, namely $sum_{k=0}^{m} binom{2m}{2k}=2^{2m-1}$(Here note that $sum_{k=0}^{m-1} binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
                      begin{eqnarray*}sum_{k=0}^{m+1} binom{2m+2}{2k}&=&sum_{k=0}^{m+1}left{ binom{2m}{2k-1}+binom{2m}{2k-2}+binom{2m}{2k}+binom{2m}{2k-1}right}\
                      &=&sum_{k=0}^{m+1} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+2sum_{k=0}^{m-1} binom{2m}{2k+1}+sum_{k=0}^{m} binom{2m}{2k}\
                      &=&4.2^{2m-1}=2^{2m+1}end{eqnarray*}                       by use of $binom{n}{k}=binom{n-1}{k-1}+binom{n-1}{k}$. Hence equality $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ is valid for $ninmathbb{N}$ which implies $(*)$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                        $endgroup$
                        – Nick
                        Jan 24 at 12:58










                      • $begingroup$
                        Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                        $endgroup$
                        – guest
                        Jan 24 at 13:41
















                      2












                      2








                      2





                      $begingroup$

                      For brevity we prove $$sum_{k=0}^{n} binom{2n}{2k} = sum_{k=0}^{n-1} binom{2n}{2k+1} = 2^{2n-1},hspace{2cm}(*)$$ for $ninmathbb{N}$. It is sufficient to show that $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ since we have
                      $$sum_{k=0}^{n} binom{2n}{2k} +sum_{k=0}^{n-1} binom{2n}{2k+1}=sum_{k=0}^{2n} binom{2n}{k} = 2^{2n}.hspace{2cm}(**)$$ Let apply induction for $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$.
                      For $n=1$ equality is true. For $n=m$ assume equality is true, namely $sum_{k=0}^{m} binom{2m}{2k}=2^{2m-1}$(Here note that $sum_{k=0}^{m-1} binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
                      begin{eqnarray*}sum_{k=0}^{m+1} binom{2m+2}{2k}&=&sum_{k=0}^{m+1}left{ binom{2m}{2k-1}+binom{2m}{2k-2}+binom{2m}{2k}+binom{2m}{2k-1}right}\
                      &=&sum_{k=0}^{m+1} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+2sum_{k=0}^{m-1} binom{2m}{2k+1}+sum_{k=0}^{m} binom{2m}{2k}\
                      &=&4.2^{2m-1}=2^{2m+1}end{eqnarray*}                       by use of $binom{n}{k}=binom{n-1}{k-1}+binom{n-1}{k}$. Hence equality $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ is valid for $ninmathbb{N}$ which implies $(*)$.






                      share|cite|improve this answer











                      $endgroup$



                      For brevity we prove $$sum_{k=0}^{n} binom{2n}{2k} = sum_{k=0}^{n-1} binom{2n}{2k+1} = 2^{2n-1},hspace{2cm}(*)$$ for $ninmathbb{N}$. It is sufficient to show that $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ since we have
                      $$sum_{k=0}^{n} binom{2n}{2k} +sum_{k=0}^{n-1} binom{2n}{2k+1}=sum_{k=0}^{2n} binom{2n}{k} = 2^{2n}.hspace{2cm}(**)$$ Let apply induction for $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$.
                      For $n=1$ equality is true. For $n=m$ assume equality is true, namely $sum_{k=0}^{m} binom{2m}{2k}=2^{2m-1}$(Here note that $sum_{k=0}^{m-1} binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
                      begin{eqnarray*}sum_{k=0}^{m+1} binom{2m+2}{2k}&=&sum_{k=0}^{m+1}left{ binom{2m}{2k-1}+binom{2m}{2k-2}+binom{2m}{2k}+binom{2m}{2k-1}right}\
                      &=&sum_{k=0}^{m+1} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+ 2sum_{k=0}^{m+1} binom{2m}{2k-1}+sum_{k=0}^{m+1} binom{2m}{2k-2}\
                      &=&sum_{k=0}^{m} binom{2m}{2k}+2sum_{k=0}^{m-1} binom{2m}{2k+1}+sum_{k=0}^{m} binom{2m}{2k}\
                      &=&4.2^{2m-1}=2^{2m+1}end{eqnarray*}                       by use of $binom{n}{k}=binom{n-1}{k-1}+binom{n-1}{k}$. Hence equality $sum_{k=0}^{n} binom{2n}{2k}=2^{2n-1}$ is valid for $ninmathbb{N}$ which implies $(*)$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 24 at 13:36

























                      answered Jan 24 at 12:14









                      guestguest

                      775416




                      775416












                      • $begingroup$
                        Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                        $endgroup$
                        – Nick
                        Jan 24 at 12:58










                      • $begingroup$
                        Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                        $endgroup$
                        – guest
                        Jan 24 at 13:41




















                      • $begingroup$
                        Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                        $endgroup$
                        – Nick
                        Jan 24 at 12:58










                      • $begingroup$
                        Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                        $endgroup$
                        – guest
                        Jan 24 at 13:41


















                      $begingroup$
                      Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                      $endgroup$
                      – Nick
                      Jan 24 at 12:58




                      $begingroup$
                      Could you please explain the transformation in more detail from $sum_{k=0}^{n/2} {nchoose2k} = sum_{k=0}^{n/2 - 1} {nchoose2k+1} = 2^{n-1}$ to $sum_{k=0}^{n} {2nchoose2k} = sum_{k=0}^{n - 1} {2nchoose2k+1} = 2^{2n-1}$
                      $endgroup$
                      – Nick
                      Jan 24 at 12:58












                      $begingroup$
                      Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                      $endgroup$
                      – guest
                      Jan 24 at 13:41






                      $begingroup$
                      Since you take $n$ as even natural number in your question, we can make transformation $n=2m$ where $minmathbb{N}$. Then, your formula transforms accordingly.
                      $endgroup$
                      – guest
                      Jan 24 at 13:41




















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