How is the partial sum of a geometric sequence calculated?

The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.

Hint: Think derivatives w.r.t. $b\,$:

$$1 + b + 2b^2 + 3b^3 + \dotsm + Nb^N=b\sum_{k=1}^{N}k\,b^{k-1}=b\Bigl(\sum_{k=0}^{N}b^k\Bigr)^{\!\!'}$$

Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $\sum_{k=0}^N x^k$ is not the same as $\sum_{i=0}^Ni*x^i$.

Try to solve the problem using the principle of induction:

Let us assume the above relation holds for $n$: $\sum_{i=0}^n ix^i=\frac{(nx-n-1)x^{n+1}+x}{(1-x)^2}\equiv=A_n$.

Now let us show it must also holds for $n+1$ and therefore for all $n\in\mathcal{N}$. Thus,

\begin{align} \sum_{i=0}^{n+1}ix^i&=\underbrace{\sum_{i=0}^{n}ix^i}_{=A_n}+(n+1)x^{n+1}\\ &=A_n+(n+1)x^{n+1}\\ &=\frac{(nx-n-1)x^{n+1}+x}{(1-x)^2}+(n+1)x^{n+1}\\ &=\frac{(nx-n-1)x^{n+1}+x}{(1-x)^2}+(n+1)x^{n+1}\frac{(1-x)^2}{(1-x)^2}\\ &=\frac{(nx-n-1)x^{n+1}+x+(n+1)x^{n+1}*(x-1)^2}{(1-x)^2}\\ &=\frac{1}{(1-x)^2}\biggl((nx-n-1)x^{n+1}+x+(n+1)x^{n+1}(1+x^2-2x)\biggr)\\ &=\frac{1}{(1-x)^2}\biggl(x^{n+2}((n+1)x-n-2)+x\biggr)\\ &=\frac{1}{(1-x)^2}\biggl(x^{(n+1)+1}((n+1)x-(n+1)-1)+x\biggr) \end{align}

So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.