Does the following hold for every $i\in \mathbb{N} $ : $\sum_{k=0}^{i} a^{k}=\frac{a^{i+1}-1}{a-1}$

Question

I know that for $ |a| < 1$ the sum $\sum_{k=0}^{\infty} a^{k} = \frac{1}{1-a}$

But I don't know this equation and whether it holds for every $i\in \mathbb{N} $

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$\sum_{k=0}^{i} a^{k}=\frac{a^{i+1}-1}{a-1}$

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Answer 1

Assuming you're looking for a proof of this identity, let us start by defining $S_i = \sum_{k=0}^{i} a^k = 1 + a + a^2 + \cdots + a^i$. We then multiply both sides by $a$ to get $aS_i = a + a^2 + a^3 + \cdots + a^{i+1}$. Then we subtract these two terms from eachother which will cancel out a lot of terms, this gives $S_i - aS_i = 1 - a^{i+1}$. Take out a factor $S_i$ on the LHS and divide by $1 - a$ to get that: $$ s = \frac{1 - a^{i+1}}{1 - a} = \sum_{k=0}^{i} a^k $$ Which can be shown to be equivalent to your definition.