How do you find a power series representation for # x^2 / ( 1 - 2x )^2#?

The idea is to relate this expression to the known power series expansion

#1/(1-x)=sum_(n=0)^oox^n#

Temporarily disregard the #x^2# and consider

#f(x)=x^2 1/(1-2x)^2#.

Take the integral of #1/(1-2x)^2#:

#intdx/(1-2x)^2#

Quick substitution:

#u=1-2x#

#du=-2dx, -1/2du=dx#

#-1/2intu^-2du=1/(2u)=1/(2(1-2x))#

Thus, knowing that differentiating this integrated expression returns the original #1/(1-2x)^2,# we can say

#f(x)=x^2d/dx1/(2(1-2x))#

Let's find the power series representation for the differentiated expression:

#f(x)=x^2d/dx1/2*1/(1-2x)#

We can easily relate #1/(1-2x)# to #1/(1-x)=sum_(n=0)^oox^n#:

#1/(1-2x)=sum_(n=0)^oo(2x)^n=sum_(n=0)^oo2^nx^n#

So,

#f(x)=x^2d/dx1/2sum_(n=0)^oo2^nx^n#

We can absorb the #1/2# in:

#f(x)=x^2d/dxsum_(n=0)^oo2^(n-1)x^n#

Differentiate the summation with respect to #x#, recalling that differentiating the summation causes the index to shift up by #1#:

#f(x)=x^2sum_(n=0)^ood/dx2^(n-1)x^n#

#f(x)=x^2sum_(n=1)^oo2^(n-1)nx^(n-1)#

Multiply in the #x^2:#

#f(x)=sum_(n=1)^oo2^(n-1)nx^(n-1+2)#

#f(x)=sum_(n=1)^oo2^(n-1)nx^(n+1)#