Prove that:
$$\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\binom{n}{k}=\frac{1}{n+1}$$
Solution
We have successively:
$$\begin{align*}\sum_{k=0}^n\frac{(-1)^k}{k+1}\binom{n}{k}&=\frac{1}{n+1}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\\&=\frac{1}{n+1}\left[\binom{n}{0}-\sum_{k=0}^{n+1} (-1)^k\binom{n+1}{k}\right]\\&=\frac{1}{n+1}\end{align*}$$
$$\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\binom{n}{k}=\frac{1}{n+1}$$
Solution
We have successively:
$$\begin{align*}\sum_{k=0}^n\frac{(-1)^k}{k+1}\binom{n}{k}&=\frac{1}{n+1}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\\&=\frac{1}{n+1}\left[\binom{n}{0}-\sum_{k=0}^{n+1} (-1)^k\binom{n+1}{k}\right]\\&=\frac{1}{n+1}\end{align*}$$
Or by invoking the binomial coefficient we have that:
ReplyDelete$$(1-x)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^n$$
Integrating the last equation from $0$ to $1$ yields the result.