How to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$?












0












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I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.






combinatorics summation permutations







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  • 1




    $begingroup$
    $$(1-1)^n+(1+1)^n=?$$
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:27






  • 1




    $begingroup$
    Is $n{{}}$ even?
    $endgroup$
    – Lord Shark the Unknown
    Nov 23 '18 at 19:27
















0












$begingroup$


I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.






combinatorics summation permutations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$(1-1)^n+(1+1)^n=?$$
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:27






  • 1




    $begingroup$
    Is $n{{}}$ even?
    $endgroup$
    – Lord Shark the Unknown
    Nov 23 '18 at 19:27














0












0








0





$begingroup$


I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.






combinatorics summation permutations







share|cite|improve this question









$endgroup$




I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.






combinatorics summation permutations




combinatorics summation permutations






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share|cite|improve this question











share|cite|improve this question




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asked Nov 23 '18 at 19:25









AlexAlex

70248




70248








  • 1




    $begingroup$
    $$(1-1)^n+(1+1)^n=?$$
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:27






  • 1




    $begingroup$
    Is $n{{}}$ even?
    $endgroup$
    – Lord Shark the Unknown
    Nov 23 '18 at 19:27














  • 1




    $begingroup$
    $$(1-1)^n+(1+1)^n=?$$
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:27






  • 1




    $begingroup$
    Is $n{{}}$ even?
    $endgroup$
    – Lord Shark the Unknown
    Nov 23 '18 at 19:27








1




1




$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27




$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27




1




1




$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27




$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27










3 Answers
3






active

oldest

votes


















2












$begingroup$

This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint



    $binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
    $sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Note that small $n$ needs special treatment (hence the smiley)
      $endgroup$
      – Hagen von Eitzen
      Nov 23 '18 at 20:07



















    0












    $begingroup$

    You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.



    The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.






          share|cite|improve this answer









          $endgroup$



          This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '18 at 19:28









          angryavianangryavian

          39.5k23280




          39.5k23280























              1












              $begingroup$

              Hint



              $binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
              $sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Note that small $n$ needs special treatment (hence the smiley)
                $endgroup$
                – Hagen von Eitzen
                Nov 23 '18 at 20:07
















              1












              $begingroup$

              Hint



              $binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
              $sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Note that small $n$ needs special treatment (hence the smiley)
                $endgroup$
                – Hagen von Eitzen
                Nov 23 '18 at 20:07














              1












              1








              1





              $begingroup$

              Hint



              $binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
              $sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.






              share|cite|improve this answer











              $endgroup$



              Hint



              $binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
              $sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 23 '18 at 19:36

























              answered Nov 23 '18 at 19:29









              Anurag AAnurag A

              25.8k12249




              25.8k12249












              • $begingroup$
                Note that small $n$ needs special treatment (hence the smiley)
                $endgroup$
                – Hagen von Eitzen
                Nov 23 '18 at 20:07


















              • $begingroup$
                Note that small $n$ needs special treatment (hence the smiley)
                $endgroup$
                – Hagen von Eitzen
                Nov 23 '18 at 20:07
















              $begingroup$
              Note that small $n$ needs special treatment (hence the smiley)
              $endgroup$
              – Hagen von Eitzen
              Nov 23 '18 at 20:07




              $begingroup$
              Note that small $n$ needs special treatment (hence the smiley)
              $endgroup$
              – Hagen von Eitzen
              Nov 23 '18 at 20:07











              0












              $begingroup$

              You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.



              The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.



                The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.



                  The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.






                  share|cite|improve this answer









                  $endgroup$



                  You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.



                  The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 '18 at 19:30









                  msmmsm

                  1,248515




                  1,248515






























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