Question

if $(x+1)(x+2)(x+3)...(x+n)=A_{0}+A_{1}x+..+A_{n}x^n$
then sum of 
$A_{1}+2A_{2}+3A_{3}+...+nA_{n}=?$

Answer 1

well just put x=1 first u get
$(1+1)(1+2)(1+3)...(1+n)=(n+1)!=A_{0}+A_{1}+..A_{n}$
as
$(x+1)(x+2)(x+3)...(x+n)=A_{0}+A_{1}x+..+A_{n}x^n$
take log on both sides
$log (x+1)+log(x+2)+...+log(x+n)=log (A_{0}+A_{1}x+..+A_{n}x^n)$
diff on both sides
$\frac{1}{1+x}+ \frac{1}{x+2}+...+\frac{1}{x+n}= \frac{A_{1}+2A_{2}x+..+nA_{n}x{n-1}}{A_{0}+A_{1}x+..+A_{n}x^n}$
again just put x=1
$1+ \frac{1}{2}+..+ \frac{1}{n+1}= \frac{A_{1}+2A_{2}+..+nA_{n}}{(n+1)!}$
therefore
$A_{1}+2A_{2}+..+nA_{n}=(n+1)!(1+ \frac{1}{2}+...+ \frac{1}{1+n} )$