This is a fun problem to solve!
First we find the series representation of the basic, 1/(1-x).
If you already know the answer, it is 1+x+x^2+x^3+x^4...., easy to remember.
If not, we can use the binomial expansion:
1/(1-x) = (1-x)^(-1) = 1+((-1)/1!)(-x)+(-1)(-2)/2!(-x)^2+(-1)(-2)(-3)/3!(-x)^3+...
which gives 1+x+x^2+x^3+x^4+...
Then 1/(1-x)^2 is just (1/(1-x))^2, or
(1+x+x^2+x^3+x^4+...)^2=
x+x^2+x^3+x^4+x^5+x^6+...
x^2+x^3+x^4+x^5+x^6+...
+x^3+x^4+x^5+x^6+...
+x^4+x^5+x^6+...
+x^5+x^6+...
....
....)
=1+2x+3x^2+4x^3+5x^4+6x^5+....
Similarly,
(1+x)/(1-x)^2 can be considered as
=1/(1+x)^2+x(1+x)^2
=1+2x+3x^2+4x^3+5x^4+6x^5+....
+x+2x^2+3x^3+4x^4+5x^5+...
=1+3x+5x^2+7x^3+9x^4+11x^5+...
Finally,
x(1+x)/(1-x)^2 can be considered as
=x*(1+x)/(1-x)^2
=x*(1+3x+5x^2+7x^3+9x^4+11x^5+...)
=x+3x^2+5x^3+7x^4+9x^5+11x^6+....+(2i-1)x^i+...
Therefore
x(1+x)/(1-x)^2 = x+3x^2+5x^3+7x^4+9x^5+11x^6+...+(2i-1)x^i+... ad infinitum
