A question on binomial theorem












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If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$






binomial-coefficients binomial-theorem







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    math.stackexchange.com/questions/437523/…
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    – lab bhattacharjee
    Jan 10 at 12:29
















2












$begingroup$


If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$






binomial-coefficients binomial-theorem







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  • $begingroup$
    math.stackexchange.com/questions/437523/…
    $endgroup$
    – lab bhattacharjee
    Jan 10 at 12:29














2












2








2





$begingroup$


If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$






binomial-coefficients binomial-theorem







share|cite|improve this question











$endgroup$




If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$






binomial-coefficients binomial-theorem




binomial-coefficients binomial-theorem






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edited Jan 10 at 12:15









darij grinberg

11.5k33168




11.5k33168










asked Jan 10 at 12:14









Supriyo HalderSupriyo Halder

658113




658113












  • $begingroup$
    math.stackexchange.com/questions/437523/…
    $endgroup$
    – lab bhattacharjee
    Jan 10 at 12:29


















  • $begingroup$
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    $endgroup$
    – lab bhattacharjee
    Jan 10 at 12:29
















$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29




$begingroup$
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– lab bhattacharjee
Jan 10 at 12:29










2 Answers
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$begingroup$

Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.




First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$




Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$






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    $begingroup$

    In other words, you are saying that
    begin{equation}
    sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
    end{equation}
    (because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).



    This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.



    (If you're looking for a hint: Induction on $n$.)






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      Evaluate
      $$int_0^1frac{1-(1-x)^n}xdx$$
      in two different ways.




      First way:
      $$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
      so
      $$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
      $$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$




      Second way: Substituting $u=1-x$,
      $$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Evaluate
        $$int_0^1frac{1-(1-x)^n}xdx$$
        in two different ways.




        First way:
        $$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
        so
        $$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
        $$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$




        Second way: Substituting $u=1-x$,
        $$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Evaluate
          $$int_0^1frac{1-(1-x)^n}xdx$$
          in two different ways.




          First way:
          $$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
          so
          $$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
          $$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$




          Second way: Substituting $u=1-x$,
          $$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$






          share|cite|improve this answer











          $endgroup$



          Evaluate
          $$int_0^1frac{1-(1-x)^n}xdx$$
          in two different ways.




          First way:
          $$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
          so
          $$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
          $$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$




          Second way: Substituting $u=1-x$,
          $$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 23:28

























          answered Jan 10 at 13:11









          bofbof

          52.6k559121




          52.6k559121























              1












              $begingroup$

              In other words, you are saying that
              begin{equation}
              sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
              end{equation}
              (because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).



              This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.



              (If you're looking for a hint: Induction on $n$.)






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In other words, you are saying that
                begin{equation}
                sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
                end{equation}
                (because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).



                This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.



                (If you're looking for a hint: Induction on $n$.)






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In other words, you are saying that
                  begin{equation}
                  sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
                  end{equation}
                  (because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).



                  This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.



                  (If you're looking for a hint: Induction on $n$.)






                  share|cite|improve this answer









                  $endgroup$



                  In other words, you are saying that
                  begin{equation}
                  sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
                  end{equation}
                  (because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).



                  This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.



                  (If you're looking for a hint: Induction on $n$.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 12:18









                  darij grinbergdarij grinberg

                  11.5k33168




                  11.5k33168






























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