A question about combinatorics: $sum_{substack{0le kle n\ ktext{ even}}}frac1{k+1}binom nk$












0












$begingroup$


enter image description here




Let $ninmathbb N$ be fixed. For $0le k le n$, et $C_k=binom nk$. Evaluate: $$sum_{substack{0le kle n\ ktext{ even}}} frac{C_k}{k+1}.$$




My attempt : $(x+1)^n=sum_{k=0}^{n}{binom{n}{k}x^k}implies frac{(x+1)^{n+1}}{n+1}+C=int (x+1)^ndx=sum_{k=0}^{n}{binom{n}{k}frac{1}{k+1}x^{k+1}}.$



Putting $x=0$ we have $C=frac{-1}{n+1}$.



so my answer is $$frac{2^{n+1}-1}{n+1}$$



is its True ?



Any hints/solution will appreciated



thanks u






combinatorics summation binomial-coefficients







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$endgroup$








  • 4




    $begingroup$
    Did you consider that $k$ is even?
    $endgroup$
    – harshit54
    Jan 13 at 17:24










  • $begingroup$
    i thinks its answer will be remain same when k is even also @harshit54
    $endgroup$
    – jasmine
    Jan 13 at 17:29








  • 1




    $begingroup$
    No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
    $endgroup$
    – harshit54
    Jan 13 at 17:31






  • 1




    $begingroup$
    When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
    $endgroup$
    – A.Γ.
    Jan 13 at 17:55
















0












$begingroup$


enter image description here




Let $ninmathbb N$ be fixed. For $0le k le n$, et $C_k=binom nk$. Evaluate: $$sum_{substack{0le kle n\ ktext{ even}}} frac{C_k}{k+1}.$$




My attempt : $(x+1)^n=sum_{k=0}^{n}{binom{n}{k}x^k}implies frac{(x+1)^{n+1}}{n+1}+C=int (x+1)^ndx=sum_{k=0}^{n}{binom{n}{k}frac{1}{k+1}x^{k+1}}.$



Putting $x=0$ we have $C=frac{-1}{n+1}$.



so my answer is $$frac{2^{n+1}-1}{n+1}$$



is its True ?



Any hints/solution will appreciated



thanks u






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Did you consider that $k$ is even?
    $endgroup$
    – harshit54
    Jan 13 at 17:24










  • $begingroup$
    i thinks its answer will be remain same when k is even also @harshit54
    $endgroup$
    – jasmine
    Jan 13 at 17:29








  • 1




    $begingroup$
    No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
    $endgroup$
    – harshit54
    Jan 13 at 17:31






  • 1




    $begingroup$
    When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
    $endgroup$
    – A.Γ.
    Jan 13 at 17:55














0












0








0





$begingroup$


enter image description here




Let $ninmathbb N$ be fixed. For $0le k le n$, et $C_k=binom nk$. Evaluate: $$sum_{substack{0le kle n\ ktext{ even}}} frac{C_k}{k+1}.$$




My attempt : $(x+1)^n=sum_{k=0}^{n}{binom{n}{k}x^k}implies frac{(x+1)^{n+1}}{n+1}+C=int (x+1)^ndx=sum_{k=0}^{n}{binom{n}{k}frac{1}{k+1}x^{k+1}}.$



Putting $x=0$ we have $C=frac{-1}{n+1}$.



so my answer is $$frac{2^{n+1}-1}{n+1}$$



is its True ?



Any hints/solution will appreciated



thanks u






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$




enter image description here




Let $ninmathbb N$ be fixed. For $0le k le n$, et $C_k=binom nk$. Evaluate: $$sum_{substack{0le kle n\ ktext{ even}}} frac{C_k}{k+1}.$$




My attempt : $(x+1)^n=sum_{k=0}^{n}{binom{n}{k}x^k}implies frac{(x+1)^{n+1}}{n+1}+C=int (x+1)^ndx=sum_{k=0}^{n}{binom{n}{k}frac{1}{k+1}x^{k+1}}.$



Putting $x=0$ we have $C=frac{-1}{n+1}$.



so my answer is $$frac{2^{n+1}-1}{n+1}$$



is its True ?



Any hints/solution will appreciated



thanks u






combinatorics summation binomial-coefficients




combinatorics summation binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 14:48









Martin Sleziak

44.9k10121274




44.9k10121274










asked Jan 13 at 17:22









jasminejasmine

1,888418




1,888418








  • 4




    $begingroup$
    Did you consider that $k$ is even?
    $endgroup$
    – harshit54
    Jan 13 at 17:24










  • $begingroup$
    i thinks its answer will be remain same when k is even also @harshit54
    $endgroup$
    – jasmine
    Jan 13 at 17:29








  • 1




    $begingroup$
    No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
    $endgroup$
    – harshit54
    Jan 13 at 17:31






  • 1




    $begingroup$
    When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
    $endgroup$
    – A.Γ.
    Jan 13 at 17:55














  • 4




    $begingroup$
    Did you consider that $k$ is even?
    $endgroup$
    – harshit54
    Jan 13 at 17:24










  • $begingroup$
    i thinks its answer will be remain same when k is even also @harshit54
    $endgroup$
    – jasmine
    Jan 13 at 17:29








  • 1




    $begingroup$
    No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
    $endgroup$
    – harshit54
    Jan 13 at 17:31






  • 1




    $begingroup$
    When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
    $endgroup$
    – A.Γ.
    Jan 13 at 17:55








4




4




$begingroup$
Did you consider that $k$ is even?
$endgroup$
– harshit54
Jan 13 at 17:24




$begingroup$
Did you consider that $k$ is even?
$endgroup$
– harshit54
Jan 13 at 17:24












$begingroup$
i thinks its answer will be remain same when k is even also @harshit54
$endgroup$
– jasmine
Jan 13 at 17:29






$begingroup$
i thinks its answer will be remain same when k is even also @harshit54
$endgroup$
– jasmine
Jan 13 at 17:29






1




1




$begingroup$
No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
$endgroup$
– harshit54
Jan 13 at 17:31




$begingroup$
No it should not. If k is even then your are summing up $C_0+C_2+C_4+C_6......$. What you have summed is $C_0+C_1+C_2+C_3......$.
$endgroup$
– harshit54
Jan 13 at 17:31




1




1




$begingroup$
When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
$endgroup$
– A.Γ.
Jan 13 at 17:55




$begingroup$
When it comes to integration: $$int_{-1}^1 x^k,dx=0qquad ktext{ odd}.$$
$endgroup$
– A.Γ.
Jan 13 at 17:55










2 Answers
2






active

oldest

votes


















1












$begingroup$

In your solution you have added all terms, whereas it should be done only for even ones:



$$sum_{substack{0leq k leq n\ktext{ even}}}frac 1{k+1}binom nk=sum_{k=0}^nfrac 1{k+1}binom nkleft[frac{x^{k+1}-(-x)^{k+1}}{2}right]_0^1\=int_0^1frac{(x+1)^n-(1-x)^n}{2}dx
=frac{2^n}{n+1}.
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint :
    Observe that
    begin{align*}
    2sum_{substack{0leq k leq n\ktext{ even}}} frac{C_k}{k+1} &= sum_{0leq kleq n} frac{C_k}{k+1} +sum_{0leq kleq n} frac{C_k}{k+1} (-1)^k
    end{align*}



    You can apply the rest of your reasoning to the two sums above to get your desired result.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      can u elaborate more ......??
      $endgroup$
      – jasmine
      Jan 13 at 17:56






    • 1




      $begingroup$
      This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
      $endgroup$
      – P. Quinton
      Jan 13 at 17:57











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    In your solution you have added all terms, whereas it should be done only for even ones:



    $$sum_{substack{0leq k leq n\ktext{ even}}}frac 1{k+1}binom nk=sum_{k=0}^nfrac 1{k+1}binom nkleft[frac{x^{k+1}-(-x)^{k+1}}{2}right]_0^1\=int_0^1frac{(x+1)^n-(1-x)^n}{2}dx
    =frac{2^n}{n+1}.
    $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      In your solution you have added all terms, whereas it should be done only for even ones:



      $$sum_{substack{0leq k leq n\ktext{ even}}}frac 1{k+1}binom nk=sum_{k=0}^nfrac 1{k+1}binom nkleft[frac{x^{k+1}-(-x)^{k+1}}{2}right]_0^1\=int_0^1frac{(x+1)^n-(1-x)^n}{2}dx
      =frac{2^n}{n+1}.
      $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        In your solution you have added all terms, whereas it should be done only for even ones:



        $$sum_{substack{0leq k leq n\ktext{ even}}}frac 1{k+1}binom nk=sum_{k=0}^nfrac 1{k+1}binom nkleft[frac{x^{k+1}-(-x)^{k+1}}{2}right]_0^1\=int_0^1frac{(x+1)^n-(1-x)^n}{2}dx
        =frac{2^n}{n+1}.
        $$






        share|cite|improve this answer











        $endgroup$



        In your solution you have added all terms, whereas it should be done only for even ones:



        $$sum_{substack{0leq k leq n\ktext{ even}}}frac 1{k+1}binom nk=sum_{k=0}^nfrac 1{k+1}binom nkleft[frac{x^{k+1}-(-x)^{k+1}}{2}right]_0^1\=int_0^1frac{(x+1)^n-(1-x)^n}{2}dx
        =frac{2^n}{n+1}.
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 13 at 18:20

























        answered Jan 13 at 17:58









        useruser

        5,47411030




        5,47411030























            1












            $begingroup$

            Hint :
            Observe that
            begin{align*}
            2sum_{substack{0leq k leq n\ktext{ even}}} frac{C_k}{k+1} &= sum_{0leq kleq n} frac{C_k}{k+1} +sum_{0leq kleq n} frac{C_k}{k+1} (-1)^k
            end{align*}



            You can apply the rest of your reasoning to the two sums above to get your desired result.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              can u elaborate more ......??
              $endgroup$
              – jasmine
              Jan 13 at 17:56






            • 1




              $begingroup$
              This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
              $endgroup$
              – P. Quinton
              Jan 13 at 17:57
















            1












            $begingroup$

            Hint :
            Observe that
            begin{align*}
            2sum_{substack{0leq k leq n\ktext{ even}}} frac{C_k}{k+1} &= sum_{0leq kleq n} frac{C_k}{k+1} +sum_{0leq kleq n} frac{C_k}{k+1} (-1)^k
            end{align*}



            You can apply the rest of your reasoning to the two sums above to get your desired result.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              can u elaborate more ......??
              $endgroup$
              – jasmine
              Jan 13 at 17:56






            • 1




              $begingroup$
              This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
              $endgroup$
              – P. Quinton
              Jan 13 at 17:57














            1












            1








            1





            $begingroup$

            Hint :
            Observe that
            begin{align*}
            2sum_{substack{0leq k leq n\ktext{ even}}} frac{C_k}{k+1} &= sum_{0leq kleq n} frac{C_k}{k+1} +sum_{0leq kleq n} frac{C_k}{k+1} (-1)^k
            end{align*}



            You can apply the rest of your reasoning to the two sums above to get your desired result.






            share|cite|improve this answer









            $endgroup$



            Hint :
            Observe that
            begin{align*}
            2sum_{substack{0leq k leq n\ktext{ even}}} frac{C_k}{k+1} &= sum_{0leq kleq n} frac{C_k}{k+1} +sum_{0leq kleq n} frac{C_k}{k+1} (-1)^k
            end{align*}



            You can apply the rest of your reasoning to the two sums above to get your desired result.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 13 at 17:50









            P. QuintonP. Quinton

            1,9061213




            1,9061213












            • $begingroup$
              can u elaborate more ......??
              $endgroup$
              – jasmine
              Jan 13 at 17:56






            • 1




              $begingroup$
              This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
              $endgroup$
              – P. Quinton
              Jan 13 at 17:57


















            • $begingroup$
              can u elaborate more ......??
              $endgroup$
              – jasmine
              Jan 13 at 17:56






            • 1




              $begingroup$
              This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
              $endgroup$
              – P. Quinton
              Jan 13 at 17:57
















            $begingroup$
            can u elaborate more ......??
            $endgroup$
            – jasmine
            Jan 13 at 17:56




            $begingroup$
            can u elaborate more ......??
            $endgroup$
            – jasmine
            Jan 13 at 17:56




            1




            1




            $begingroup$
            This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
            $endgroup$
            – P. Quinton
            Jan 13 at 17:57




            $begingroup$
            This tricks permits to remove the "even" condition. having done that, these can be handled exactly as you did, actually you already did the first one and the second one is the same with $x=-1$
            $endgroup$
            – P. Quinton
            Jan 13 at 17:57


















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