# 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

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HINT $\$ It's a geometric series

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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.

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... and where there are $n$ terms. (Be careful, as there are $n+1$ terms in the question) – Henry Feb 26 '11 at 9:33

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

$$S=1+1+1^{2}+1^{3}+\ldots +1^{n}=1+n.$$

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Why consider $(1+x)^0$ separately? – Yuval Filmus Feb 26 '11 at 18:23
Yuval Filmus, You are right! There's no need. The first term becomes $1$ and the ratio remains $(1+x)$ – Américo Tavares Feb 26 '11 at 18:31