<s>[INST] If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function 
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$ 
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
 [/INST] \begin{eqnarray*}
\frac{1}{x(x+1)}&=&\frac{1}{x}-\frac{1}{x+1} \\
\frac{2}{x(x+1)(x+2)}&=&\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2} \\
\frac{6}{x(x+1)(x+2)(x+3)}&=&\frac{2}{x(x+1)(x+2)}-\frac{2}{(x+1)(x+2)(x+3)}\\&=&\frac{1}{x}-\frac{3}{x+1}+\frac{3}{x+2}- \frac{1}{x+3}\\
\cdots
\end{eqnarray*}
$A_7= \color{red}{\binom{n}{7}}$.
</s>