Answers

The Brainliest Answer!
2014-04-10T20:22:31+05:30
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}  )

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