Formula for sum of series $1+(1+x)+(1+x)^2+.... +(1+x)^n$ What is the formula to get sum of series 
$$1+(1+x)+(1+x)^2+.... +(1+x)^n,$$ 
where $n$ is Integer and $x$ is Rational

Simplified after the comment below by Yuval Filmus. 
$$S=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}$$ 
is the sum of a  geometric progression with ratio $1+x$ and $n+1$ terms, the first of which is $1$. The usual way to derive $S$ is to multiply it by the ratio
$$(1+x)S=(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}+(1+x)^{n+1}$$
and subtract from $S$:
$$\begin{eqnarray*}
S-(1+x)S &=&1+\left( (1+x)-(1+x)\right) +\left(
(1+x)^{2}-(1+x)^{2}\right)  \\
&&+\ldots +\left( (1+x)^{n}-(1+x)^{n}\right) -(1+x)^{n+1} \\
&=&1-(1+x)^{n+1}.
\end{eqnarray*}$$
Solving for $S$, we get
$$S=\frac{(1+x)^{n+1}-1}{x}\qquad (x\neq 0).$$
For $x=0$ the sum of the series is
$$S=1+1+1^{2}+1^{3}+\ldots +1^{n}=1+n.$$ 
