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)=\frac1{1-x}=\displaystyle\sum_(n=0)^\infty x^n`

with -1 < x < 1. Differentiating gives

`f'(x)=\frac1{(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=(1+x)/((1-x)^2)`