If $\left(x\right)<1,$ and $r$th term of a series is $1 + x$$+x^2+\dots +x^{r-1},$ then sum to $n$ terms of the series is

$t_r=1+x+x^2+\dots x^{r-1}=\frac{1-x^r}{1-x}$

Now, $\sum _{r=1}^n t_r=\sum _{r=1}^n \left(\frac{1-x^r}{1-x}\right)$

$=\frac{n}{1-x}-\frac{1}{1-x}\sum _{r=1}^n x^r$

$\begin{array}{l}=\frac{n}{1-x}-\frac{1}{1-x}\frac{x\left(1-x^n\right)}{1-x}\\ =\frac{1}{(1-x{)}^2}\left[n-nx-x+x^{n+1}\right]\\ =\frac{1}{(1-x{)}^2}\left[n-(n+1)x+x^{n+1}\right]\end{array}$