TITLE: Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$
QUESTION [5 upvotes]: Deduce that $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\dfrac{1}{n+2}$

We know $\dfrac{1}{n+2}=\displaystyle \int_{0}^{1}t^{n+2-1}dt$
Now $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)(k+2)}\bigg)=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}-\dfrac{1}{k+2}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}\bigg)-\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+2)}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\int_{0}^1 t^{k+1-1}dt \space - \space \displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k \int_{0}^1 t^{k+2-1}dt$
$=\displaystyle \int_{0}^1 t^{1-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt- \int_{0}^1 t^{2-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt$
$=\displaystyle  \int_{0}^1 t^{1-1}(1-t)^{n}dt \space - \space  \int_{0}^1 t^{2-1}(1-t)^{n}dt$
$=\displaystyle \large \beta(1,n+1)$-$\displaystyle \large\beta(2,n+1)$
$=\dfrac{\Gamma(1)\Gamma(n+1)}{\Gamma(1+n+1)}-\dfrac{\Gamma(2)\Gamma(n+1)}{\Gamma(2+n+1)}= \underbrace{\dfrac{1}{n+2}}_{\text{which is exactly what I  proved}}$ 
PS @Chappers Thankyou all users for correcting one nasty mistake.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{1 \over \pars{k + 1}\pars{k + 2}} & =
\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}
\int_{0}^{1}\pars{t^{k} - t^{k + 1}}\dd t
\\[5mm] & =
\int_{0}^{1}\sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k}\,\dd t -
\int_{0}^{1}t\sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k}\,\dd t
\\[5mm] & =
\int_{0}^{1}\pars{1 - t}^{n}\,\dd t -
\int_{0}^{1}t\pars{1 - t}^{n}\,\dd t =
\int_{0}^{1}t^{n + 1}\,\dd t
\\[5mm] & =
\bbx{1 \over n + 2}
\end{align}