<s>[INST] Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$ 
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.
 [/INST] $$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2} = \int_{0}^{1} x^k (1-x)\,dx $$
hence
$$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)} &=& \int_{0}^{1}(1-x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\ &=& \int_{0}^{1}(1-x)^{n+1}\,dx = \int_{0}^{1} x^{n+1}\,dx = \frac{1}{n+2}.\end{eqnarray*}$$
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