Find $$S=\sum_{r=1}^{3n-1}{ (-1)^{r-1}r\over{3n \choose r}},~ \text{if $n$ is even}$$

The answer given to me is ${3n\over3n+2}$ , the main problem I am facing is that the binomial coefficients are in the denominator, and so I can not use any of the usual techniques I used to use , like using integration or different on any binomial series. Although I tried by rewriting the sum in reverse order and then adding it to the original expression, this gave me :

$$2S=3n\sum_{r=1}^{3n-1}{ (-1)^{r-1}\over{3n \choose r}}$$

This is sure simpler than the original problem but still no good , I could not figure out a way to solve the rest .

Could someone please help me in solving thi problem ?

Thanks !

  • $\begingroup$ Start by rewriting the reciprocal of binomial coefficients as a Beta function $$\frac1{\binom{n}{k}}=(n+1)B(k+1,n-k+1)=(n+1)\int_0^1t^k(1-t)^{n-k}\,\mathrm{d}t.$$ $\endgroup$ Jul 25, 2020 at 15:05
  • $\begingroup$ It also helps to use$$\sum_{r=1}^{3n-1}rx^r=\frac{x-3nx^{3n}+(3n-1)x^{3n+1}}{(1-x)^2}$$(which you can prove by induction on $n\ge1$), where $x$ is a function of @user10354138's $t$ you'll guess as you work through it. $\endgroup$
    – J.G.
    Jul 25, 2020 at 15:26
  • $\begingroup$ You may see math.stackexchange.com/questions/3251604/… $\endgroup$
    – Z Ahmed
    Jul 25, 2020 at 16:10

1 Answer 1


Let $$S=\sum_{r=1}^{3n-1} \frac{(-1)^{r-1} ~r}{3n \choose r}$$ Change $r\to 3n-r$, then $$S=\sum_{r=1}^{3n-1} \frac{(-1)^{3n-r-1}~(3n-r)}{3n \choose 3n-r}= (-1)^{3n}\sum_{s=1}^{3n-1}\frac{(-1)^{r-1}(3n-r)}{{3n \choose s}}$$ $$\implies [1+(-1)^{3n}]S=3n(-1)^{3n-1}\sum_{r=1}^{3n-1}\frac{(-1)^r}{{3n \choose r}}=3n(-1)^{3n-1}\left( \sum_{r=0}^{3n} \frac{(-1)^r}{{3n \choose r}}-(1+(-1)^{3n})\right)$$ Next using Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ We get $$[1+(-1)^{3n}] S=3n(-1)^{3n-1} [1+(-1)^{3n}]\frac{3n+1}{3n+2}+3n[1+(-1)^{3n}]$$ If $3n$ is even, then $$S=3n \left(1-\frac{3n+1}{3n+2}\right)=\frac{3n}{3n+2},~\text{only if}~n ~\text{is even}$$

For odd $n$ the sum is (perhaps) not doable by hand in a closed form. One may check that $S(n=1)=-1/3, S(n=3)=-7/9.$


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