Imagine I have a series such as:
$$(1 + n)^0 + (1 + n)^1 + (1 + n)^2 + (1 + n)^3 + ... (1+n)^x$$
Is there a simple way to summarize this as a function of $n$ and $x$?
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$= \sum_{i=0}^x (1+n)^i = \sum_{i=0}^x \sum_{k=0}^i \binom{i}{k}n^k$ – Zubzub Aug 19 '16 at 12:27
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This is a geometric progression. The general formula for a geometric progression is given by $$ \sum_{k=1}^{n} ar^{k-1} = \frac{a(1-r^n)}{1-r} $$ provided that $r\neq1$. In our case, $a=1$ and $r=n+1$. Hence, $$ \sum_{k=1}^{x+1}(1+n)^{k-1}=\frac{(1+n)^{x+1}-1}{n}. $$
Cm7F7Bb
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$$I(n,x)=\color{red}{1} + (1 + n)^1 + (1 + n)^2 + (1 + n)^3 +\cdots+ (1+n)^x$$ $$(1+n)I(n,x)=(1 + n)^1 + (1 + n)^2 + (1 + n)^3 + (1 + n)^4 + \cdots +\color{red}{(1+n)^{x+1}}$$ we have $$\underbrace{(1+n)I(n,x)-I(n,x)}_{nI(n,x)}=(1+n)^{x+1}-1$$ therefore $$I(n,x)=\frac{(1+n)^{x+1}-1}{n}$$
Behrouz Maleki
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