Find $\sum_{n=1}^{\infty} n\left(2n-1\right)x^{2n}$

Question

Question: Find $$\sum_{n=1}^{\infty} n\left(2n-1\right)x^{2n}.$$

My Approach:$$ \lim_{n\rightarrow\infty} |n\left(2n-1\right)|^{\frac{1}{n}} =1 \Longrightarrow \text{Radius of convergence} = 1. $$

I don't know how to find the sum. Any Hint or help will be highly appreciated.

I Solved It,Thanks To Przemysław Scherwentke

$\sum_{n=1}^{k}$$x^{2n}$=$\frac{x^{2}\left[1-\left(x^{2}\right)^{k}\right]}{1-x}$

$\Rightarrow$$\sum_{n=1}^{\infty}$$x^{2n}$=$\frac{x^{2}}{1-x}$

Differentiating twice we get the required result .

Answer 1

Take $f(x)=\sum_{n=1}^{\infty}n(2n-1)x^{2n}$ therefore:$$g(x)=\int \frac{f(x)}{x^2}dx=C_1+\sum_{n=1}^{\infty}nx^{2n-1}\\g_1(x)=\sum_{n=1}^{\infty}nx^{2n-1}\\h(x)=\int \frac{g_1(x)}{x}dx^2=C_2+\sum_{n=1}^{\infty}x^{2n}=\frac{1}{1-x^2}-1+C_2$$Now by successive differentiation we obtain:$$g_1(x)=x\frac{dh(x)}{d(x^2)}=\frac{x}{(1-x^2)^2}\to g(x)=\frac{x}{(1-x^2)^2}+C_1$$and $$f(x)=x^2\frac{dg}{dx}=\frac{-3x^6+2x^4+x^2}{x^4-2x^2+1}$$ and finally:$$\sum_{n=1}^{\infty}n(2n-1)x^{2n}=x^2\frac{1+3x^2}{1-x^2}$$

Answer 2

HINT: $(x^{2n})''=2n\cdot(2n-1)x^{2n-2}$