Geometric summation proof, not calculus

Question

I am trying to take the expression $$T=\sum_{k=1}^nkx^k$$ and make it into a "simpler expression." I have an example similar to it where i am finding $$\sum_{k=1}^nx^k$$ where the answer is $$S_0 = {1-x^n \over 1-x}$$ and I am supposed to use that in my solution, so I solved for$$T - xT = \sum_{k=1}^nkx^k - x\sum_{k=1}^nkx^x$$ and I have $$T = {S_0 - 1 - nx^{n+1} \over (1-x) }$$ which when I plug in $S_0$, I get $$T = {nx^{n+2} - nx^{n+1} - x^n + x \over (1-x)^2}$$ From what I can find online though, this is supposed to equal $${x \over (1-x)^2}$$ and I cannot figure out how to simplify it to that. Thank you for any help!

Answer 1

Your result is corret.

The expression you found "online" is the limit as $n\to \infty$ if $|x|<1$.

Answer 2

HINT: $T=x\sum_{k=0}^{n}kx^{k-1}$ and now you can use this fact: $$\sum_{k=0}^{n}kx^{k-1}=(\sum_{k=0}^{n}x^{k})'=(\frac{1-x^{n+1}}{1-x})'$$