I managed to get
$$\sum_{k=0}^n \binom{n+1}{k+1}(-1)^k$$
on the left side, but I don't know how to proceed from here.
thanks in advance.
Asked
Active
Viewed 149 times
3
Martin Sleziak
- 53,687
Omer.S
- 43
-
Duplicate: How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? Found using Approach0. – Martin Sleziak Jan 07 '17 at 22:15
2 Answers
4
Expand $(1-1)^{n+1}$ by using the Binomial Theorem: $$0=(1-1)^{n+1}=1-\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k}=1-(n+1)\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{k+1}.$$
Robert Z
- 145,942
-
Thanks alot! but I'm not sure how you get the "1-" before the sum? thanks again! – Omer.S Jan 07 '17 at 21:00
-
@Omer.S Because $(1-1)^{n+1}=\binom{n+1}{0}+\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k+1}$. – Robert Z Jan 07 '17 at 21:03
-
I understand that $\ (1-1)^{n+1} = \sum_{k=0}^{n+1} {n+1\choose k+1} (-1)^{k+1}$ but why $\ {n+1\choose 0}$? – Omer.S Jan 07 '17 at 21:25
-
1@Omer.S $$(1-1)^{n+1}=\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^{j}=\binom{n+1}{0}+\sum_{j=1}^{n+1}\binom{n+1}{j}(-1)^{j}\=\binom{n+1}{0}+\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k+1}.$$ – Robert Z Jan 07 '17 at 21:28
3
$$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}&=&\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\&=&\int_{0}^{1}(1-x)^n\,dx\\&=&\int_{0}^{1}z^n\,dz = \color{red}{\frac{1}{n+1}}.\end{eqnarray*}$$
Jack D'Aurizio
- 353,855