<s>[INST] Proof of $\sum_{k=0}^{n}{(-1)^k{n \choose k}\frac{1}{(k+1)(k+2)}}=\frac{1}{n+2}$ 
Deduce that
  $$
{n \choose 0}\frac{1}{1\cdot 2}-{n \choose 1}\frac{1}{2\cdot 3}+{n \choose 2}\frac{1}{3\cdot4}+\cdots+{n \choose n}\frac{1}{(n+1)\cdot(n+2)}=\frac{1}{n+2}
$$

Is induction a viable method to solve this problem? Or is there some other method which might be simpler?
 [/INST] Observe that the sum is
$$\frac{1}{(n+1)(n+2)}
\sum_{k=0}^n {n+2\choose k+2} (-1)^{k+2}$$
which is
$$\frac{1}{(n+1)(n+2)}
\sum_{k=2}^{n+2} {n+2\choose k} (-1)^{k}
\\ = \frac{1}{(n+1)(n+2)}
\left(-{n+2\choose 0} (-1)^0 - {n+2\choose 1} (-1)^1\right)
\\ = \frac{1}{(n+1)(n+2)}
\left(-1 + n+2\right) = \frac{1}{n+2}.$$
</s>