Question

14. 2057 Q.No. 1(a) Sum to infinity:1 + 3x + 5x2 + 7x3 +... (-1<x<1).​

Answer 1

The sum appears to be

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n$

I'll assume you want to find out what function this sum converges to.

Let

$f(x)=\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

with -1 < x < 1. Differentiating gives

$f'(x)=\dfrac1{(1-x)^2}=\displaystyle\sum_{n=0}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty(n+1)x^n$

So we have

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n=f'(x)+xf'(x)$

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n=\frac{1+x}{(1-x)^2}$