Finding sum of a power series explicitly

Question

I was wondering how to find the sum of the power series $$\sum_{k=0}^\infty {2^k{x^k}-2k}{x^k}$$

Answer 1

For the first, you can use the technique here twice to get the $k^2$. We have $$\sum_{k=0}^\infty x^k=\frac 1{1-x}\\ \frac d{dx}\sum_{k=0}^\infty x^k=\frac d{dx}\frac 1{1-x}\\ \sum_{k=0}^\infty xkx^{k-1}=x\frac 1{(1-x)^2}\\ \frac d{dx}\sum_{k=0}^\infty xkx^{k-1}=\frac{1+x}{(1-x)^3}\\ \sum_{k=0}^\infty xk^2x^{k-1}=\frac{x+x^2}{(1-x)^3}$$

For the second, now that it has been updated, you can use the second line.

Answer 2

If $|z|<1$, then $$\sum_{k=0}^{\infty} z^k =\frac{1}{1-z} ~~~(1)$$ D.w.r.t. $z$ $$\sum_{k=1}^{\infty} k z^k =\frac{z}{(1-z)^2}~~~~(2).$$ So $$S=\sum_{k=0}^{\infty} [(2x)^k -2 k x^k]$$ By (1) and (2) we get $$S=\frac{1}{1-2x}-\frac{2x}{(1-x)^2}=\frac{5x^2-4x+1}{(1-2x)(1-x)^2}, ~if~ |x|<\frac{1}{2}.$$