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
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
HINT $\ $ It's a geometric series
It's a geometric series with common ration $(1+x)$. The sum for a geometric series is given by $$ S = a \frac{ r^{n}-1}{r-1}$$ where $a$ is the first term and $r$ is the common ratio.
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
$$S=1+1+1^{2}+1^{3}+\ldots +1^{n}=1+n.$$