Find a power series representation for the function x (1+x) / (1-x) ^2

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