TITLE: Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$
QUESTION [15 upvotes]: Various examinations ask students to prove that $$\frac{1}{{n \choose k}}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx ~~~~~(1)$$ by evaluating  the integral $\int_{0}^{1}
(tx+1-x)^n~dt$ two ways. When I came across (1), I could prove that $$\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.~~~~(2)$$ as below:  $$S_n=\frac{1}{n+1}\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=\int_{0}^1 \sum_{k=0}^{n} (-1)^k  x^k (1-x)^{n-k} dx =  \int_{0}^{1}  (1-x)^n \sum_{k=0}^{n}\left(\frac{-x}{1-x} \right)^k dx =\int_{0}^{1}(1-x)^n \frac{ \left( \frac{-x}{1-x}\right)^{n+1}-1}{\left(\frac{-x}{1-x}\right)-1}dx$$
$$=\int_{0}^{1}[(1-x)^{n+1}-(-x)^{n+1}] dx =\frac{1+(-1)^n}{n+2}.~~~~(3)$$
Therefore, $$S_{2m+1}=0 ~~~~(4) ~~~ \mbox{and}~~~ S_{2m}= \frac{2m+1}{m+1}.~~~~(5)$$
Now the question is: Can one prove (2) in some other way(s)?

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\bbox[15px,#ffd]{\sum_{k = 0}^{n}{\pars{-1}^{k} \over {n\choose k}} =
\bracks{1 + \pars{-1}^{n}}{n + 1 \over n + 2}}:\ {\Large ?}}$

\begin{align}
\sum_{k = 0}^{n}{\pars{-1}^{k} \over {n\choose k}} & =
\sum_{k = 0}^{n}\pars{-1}^{k}\,{k!\,\pars{n - k}! \over n!}
\\[5mm] & =
\pars{n + 1}\sum_{k = 0}^{n}\pars{-1}^{k}\,
{\Gamma\pars{k + 1}\Gamma\pars{n - k + 1} \over \Gamma\pars{n + 2}}
\\[5mm] & =
\pars{n + 1}\sum_{k = 0}^{n}\pars{-1}^{k}\,
\int_{0}^{1}t^{k}\pars{1 - t}^{n - k}\,\dd t
\\[5mm] & =
\pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n}
\sum_{k = 0}^{n}\pars{-\,{t \over 1 - t}}^{k}\,\dd t
\\[5mm] & =
\pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n}\,
{\bracks{-t/\pars{1 - t}}^{n + 1} - 1 \over
-t/\pars{1 - t} - 1}\,\dd t
\\[5mm] & =
\pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n + 1}\,
\bracks{1 - \pars{-\,{t \over 1 - t}}^{n + 1}}\,\dd t
\\[5mm] & =
\pars{n + 1}\int_{0}^{1}
\bracks{\pars{1 - t}^{n + 1} + \pars{-1}^{n}\, t^{n + 1}}\,\dd t
\\[5mm] & =
\pars{n + 1}
\bracks{-\,{\pars{1 - t}^{n + 2} \over n + 2}
+ \pars{-1}^{n}\, {t^{n + 2} \over n + 2}}_{\ 0}^{\ 1}
\\[5mm] & =
\pars{n + 1}
\bracks{{\pars{-1}^{n} \over n + 2} + {1 \over n + 2}}
\\[5mm] & =
\bbx{\bracks{1 + \pars{-1}^{n}}{n + 1 \over n + 2}} \\ &
\end{align}