Consider the sum
$$\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}.$$
I know the sum is zero when $n$ is odd (pretty simple).
The sum is $2-\frac{2}{2 + n}$ when $n$ is even.
Can somebody provide a proof in the even case? Thanks
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4if $n$ is odd then $$sum=\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}=\sum_{r=0}^{[n/2]} (\frac{(-1)^r}{{n \choose r}}+\frac{(-1)^{n-r}}{{n \choose {n-r}}})=\sum_{r=0}^{[n/2]} (\frac{(-1)^r}{{n \choose r}}-\frac{(-1)^{r}}{{n \choose {r}}})=\sum_{r=0}^{[n/2]} 0=0$$ – nadia-liza May 08 '14 at 21:52
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Start by writing ${1/ {n\choose r}}=(n+1)\int_0^1 u^r(1-u)^{n-r}\,du.$ Multiplying by $(-1)^r$ and adding over $r$ gives \begin{eqnarray*} \sum_{r=0}^n (-1)^r/{n\choose r} &=&(n+1)\int_0^1 \sum_{r=0}^n (-u)^r(1-u)^{n-r}\,du\\ &=&(n+1)\int_0^1 [(1-u)^{n+1}+(-1)^nu^{n+1}]\,du\\ &=&{n+1\over n+2}(1+(-1)^n). \end{eqnarray*}
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how you remember that $ 1/{{n\choose r}}=(n+1)\int_0^1 u^r(1-u)^{n-r},du $? – lele May 08 '14 at 23:44
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