Value of $\sum\limits_{n=0}^{\infty}(n+1)(0.8x)^n$

Question

I know $\sum\limits_{n=0}^{\infty}(n+1)x^n = \frac{1}{(1-x)^2}$ but does changing to $\sum\limits_{n=0}^{\infty}(n+1)(0.8x)^n$ make a difference? wolfram seems to be giving me a much more complicated result.

Answer 1

No, the result should be just as simple. If you make the substitution $y=0.8x$ you can apply your earlier result. This works because $x$ is a constant, not a variable.