$1+4x+9x^2+...=(1+x+x^2+...)+(3x+3x^2+...)+(5x^2+5x^3+...)+...=\frac{1}{1-x}+\frac{3x}{1-x}+\frac{5x^2}{1-x}+...=\frac{1+3x+5x^2+...}{1-x}=\frac{(1+x+x^2+...)+(2x+2x^2+...)+(2x^2+2x^3+...)+...}{1-x}=\frac{\frac{1}{1-x}+\frac{2x}{1-x}+\frac{2x^2}{1-x}+...}{1-x}=\frac{\frac{1}{1-x}+\frac{2x}{(1-x)^2}}{1-x}=\frac{\frac{1+x}{(1-x)^2}}{1-x}=\boxed{\frac{1+x}{(1-x)^3}}$

kootrapali wrote: $1+4x+9x^2+...=(1+x+x^2+...)+(3x+3x^2+...)+(5x^2+5x^3+...)+...=\frac{1}{1-x}+\frac{3x}{1-x}+\frac{5x^2}{1-x}+...=\frac{1+3x+5x^2+...}{1-x}=\frac{(1+x+x^2+...)+(2x+2x^2+...)+(2x^2+2x^3+...)+...}{1-x}=\frac{\frac{1}{1-x}+\frac{2x}{1-x}+\frac{2x^2}{1-x}+...}{1-x}=\frac{\frac{1}{1-x}+\frac{2x}{(1-x)^2}}{1-x}=\frac{\frac{1+x}{(1-x)^2}}{1-x}=\boxed{\frac{1+x}{(1-x)^3}}$ how did you get from $(1+x+x^2+...)+(3x+3x^2+...)+(5x^2+5x^3+...)+...$ to $\frac{1}{1-x}+\frac{3x}{1-x}+\frac{5x^2}{1-x}+...?$

Geometric Series formula since $|x|<1$ right?

Geometric series formula. Sniped

o didn't see $|x|<1$ lol.

Summation with polynomial $p(k)=k^2,\Delta p(k)=2k+1,\Delta^2 p(k)=2$ $f(n)=\frac{n^2}{q-1}-\frac{2n+1}{(q-1)^2}+\frac{2}{(q-1)^3}q$ $\sum_{k=1}^n k^2 q^{k-1}=(\frac{n^2}{q-1}-\frac{2n+1}{(q-1)^2}+\frac{2}{(q-1)^3}q)q^n+\frac{1}{(q-1)^2}-\frac{2}{(q-1)^3}q$ When $|q|<1$, $\sum_{k=1}^{\infty} k^2 q^{k-1}=\frac{1}{(q-1)^2}-\frac{2}{(q-1)^3}q=\frac{1+q}{(1-q)^3}$

Motivated by the fact the expression looks a bit like the differentiation of a polynomial, it is easy to deduce that it is equivalent to $\frac{d}{dx}(x ( \frac{d}{dx}(x+x^2+x^3+x^4 ...)))$ $\frac{d}{dx}(x (\frac{ d}{dx}(x/1-x))))$ $=\frac{1+x}{(1-x)^3}$