Question: How to find this sum ?
$$\sum\_{i=0}^{n-1}\sum\_{j=0}^{i}x^i$$
without knowing this sum:
$$\sum\_{i=1}^{n}ix^{i-1} = \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$

Answer: 
A variation based upon changing the order of summation.

> 
> We obtain
> \begin{align\*}
> \color{blue}{\sum\_{i=0}^{n-1}\sum\_{j=0}^ix^i}&=\sum\_{\color{blue}{0\leq j\leq i\leq n-1}}x^i=\sum\_{j=0}^{n-1}\sum\_{i=j}^{n-1}x^{i}\tag{1}\\
> &=\sum\_{j=0}^{n-1}\sum\_{i=0}^{n-1-j}x^{i+j}\tag{2}\\
> &=\sum\_{j=0}^{n-1}x^j\frac{1-x^{n-j}}{1-x}\tag{3}\\
> &=\frac{1}{1-x}\sum\_{j=0}^{n-1}x^j-\frac{x^n}{1-x}\sum\_{j=0}^{n-1}1\tag{4}\\
> &=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}\tag{5}\\
> &\,\,\color{blue}{=\frac{nx^{n+1}-(n+1)x^n+1}{(1-x)^2}}
> \end{align\*}
> 
> 
> 

*Comment:*

* In (1) we change the order of summation.
* In (2) we shift the index to start with $i=0$.
* In (3) we factor out $x^j$ and apply the *[geometric series formula](https://en.wikipedia.org/wiki/Geometric_series#Formula)*.
* In (4) we multiply out.
* In (5) we apply the geometric series formula again.
**Hint**:

This sum is
\begin{align}
\sum\_{i=0}^{n-1}\sum\_{j=0}^{i}x^i&=\sum\_{i=0}^{n-1}(i+1)x^i=\sum\_{i=0}^{n-1}(x^{i+1})'\\
&=\biggl(\sum\_{i=1}^{n} x^{i}\biggr)'=\dotsm
\end{align}