<s>[INST] Binomial Theorem expansion and proving an interesting identity? In the identity 
$$\frac {n!}{x(x+1)(x+2)...(x+n)}  = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \;  ... \;{(-1)^n}\; ^{n}C_n\frac1{(n+1)(n+2)}\;=\frac1{(n+2)}$$
So I have to tried to use the binomial theorem on,
$(b-a)^n$, and then multiplied both sides by $a^{x-1}$.
 Now I integrated both sides with respect to $a$. This gives me the binomially expanded side as same as the right hand side of the identity that we have to prove when I substitute $a=1$. I dont know how to integrate $a^{x-1}\;(b-a)^n$ with respect to $a$.  I am not able to prove the identity and solve the deduction. Can someone please help me out ? Thanks a lot.
 [/INST] 
We obtain
  \begin{align*}
\frac{n!}{x(x+1)\cdots(x+n)}&=\sum_{k=0}^n\frac{A_k}{x+k}\\
n!&=\sum_{k=0}^nA_k\frac{x(x+1)\cdots(x+n)}{x+k}\tag{1}
\end{align*}
  Substituting   $x=-j,0\leq j\leq n$   in (1) we get
  \begin{align*}
n!&=A_j(-j)(-j+1)\cdots   (-1)\cdots1\cdot   2\cdots(n-j)\\
&=A_j(-1)^j  j!(n-j)!\\
\end{align*}
  We get
  \begin{align*}
\color{blue}{A_j}&=(-1)^j\frac{n!}{j!(n-j)!}\\
&\,\,\color{blue}{=(-1)^j\binom{n}{j}\qquad\qquad 0\leq j\leq n}\tag{2}
\end{align*}
  and the claim follows.

Using (1) and (2) we want to show $\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$.

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+1}-\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+2}\\
&=\frac{n!}{1\cdot 2\cdots (n+1)}-\frac{n!}{2\cdot 3\cdots (n+2)}\tag{3}\\
&=\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}\\
&\,\,\color{blue}{=\frac{1}{n+2}}
\end{align*}
and the claim  follows.

In (3) we apply (1) and (2) with $x=1$ and $x=2$.
</s>