find a power series representation
f(x)=(x^2) \ (1-2x)^2
how do i do this??
Remember that, $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $\displaystyle |x| < 1$.
Thus, $\displaystyle \left( \sum_{n=0}^{\infty} x^n \right)' = \left( \frac{1}{1-x} \right)' \implies \sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$ for $\displaystyle |x| < 1$
This gives us,
$\displaystyle \sum_{n=1}^{\infty} nx^{n+1} = \frac{x^2}{(1-x)^2}, |x|<1$
Thus,
$\displaystyle \sum_{n=1}^{\infty} n(2x)^{n+1} = \frac{(2x)^2}{(1-2x)^2} \implies \sum_{n=1}^{\infty} n 2^{n-1} x^{n+1} = \frac{x^2}{(1-2x)^2}, ~ |x| < \tfrac{1}{2}$