We know that the geometric series

1 + x +`x^(2)` + ... = `sum_(n=0)^oo` `x^(n)` = `(1)/(1-x)`

converges for all |x| < 1. *Use this series to find a power series expansion for the following* (Hint: a power series can be integrated or differentiated term-by-term):

h(x) = `(x)/(1-x)^2`

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Let `h(x)=int{(1+x)/(1-x)^3}dx`

or, `h(x)=int{(1+x)(1-x)^-3}dx` (1).

Expanding `(1-x)^-3` by binomial expansion we get

`(1-x)^-3=1+3x+6x^2+10x^3+.......` (2)

Now using equation (2) , equation (1) can be written as

`h(x)=int(1+x)(1+3x+6x^2+10x^3+.......)dx`

`=int(1+4x+9x^2+16x^3+........)dx`

`=x+4x^2/2+9x^3/3+16x^4/4+.........`

`=x+2x^2+3x^3+4x^4+...........`

`=sum_(n=1)^oonx^n` .

By our question `h(x)=x/(1-x)^2` .

So, `x/(1-x)^2=sum_(n=1)^oonx^n=x+2x^2+3x^3+4x^4+........` .

Answer.

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