Answer
\[
\begin{array}{l}
S-r S=a\left(1+r+r^{2}+\ldots+r^{n}\right)-a\left(r+r^{2}+\ldots+r^{n}+r^{n+1}\right)=a-a r^{n+1} \\
S-r S=a-a r^{n+1} \Rightarrow S(1-r)=-a r^{1+n} +a \Rightarrow S=\frac{-a r^{1+n}+a}{-r+1}
\end{array}
\]
Work Step by Step
We have $S=\sum_{k=0}^{n} a r^{k}$
\[
\begin{aligned}
-r S+S &=\sum_{k=0}^{n} -r+a r^{k} \sum_{k=0}^{n} a r^{k} \\
&=-\sum_{k=0}^{n} a r^{1+k} +\sum_{k=0}^{n} a r^{k}\\
&=\left(1+r+r^{2}+\ldots+r^{n}\right)a-\left(r+r^{2}+\ldots+r^{n}+r^{n+1}\right)a \\
&=a\left(1+r+r^{2}+\ldots+r^{n}-r-r^{2}-\ldots-r^{n}-r^{n+1}\right) \\
&=a\left(-r^{1+n}+1\right)a \\
&=\overline{-a r^{1+n}+a} \\
& \therefore -r S+S=-a r^{1+n}+a
\end{aligned}
\]
\[
\because -r S+S=-a r^{1+n}+a
\]
\[
\begin{array}{l}
\Rightarrow (-r+1)S=-a r^{1+n}+a \\
\Rightarrow S=\frac{-a r^{1+n}+a}{-r+1}
\end{array}
\]
