# confused about geometric series closed formula

I have found many definitions of geometric series, for a partial sum from 0 to n-1 $$\sum_{0}^{n-1} ar^{x} = a \frac{1-r^{k}}{1-r}.$$ But the context I need to use it in, requires the sum to be from 1 to n. Is there an easy way to rewrite the closed form for this?

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Yes: replace a by a/r or by ar. –  Did Oct 5 '11 at 5:03

Answer #1: Multiply both sides by $r$ to change $a$ to $ar$, and then make the change of variables $y=x+1$ in the sum - that converts the left-hand side to what you want, while the right-hand side is multiplied by $r$.
Answer #2: It's easier to remember how geometric series identities are proved than to memorize many special cases. The denominator is always $1-r$, and to figure out what the numerator is, just multiply the series by $1-r$ and work out the resulting telescoping sum. For example, $$(1-r) \sum_{x=1}^n ar^x = (1-r)ar + (1-r)ar^2 + \cdots + (1-r)ar^n$$ $${}= ar + (-ar^2+ar^2) + (-ar^3+ar^3) + \cdots + (-ar^n + ar^n) - ar^{n-1} = ar - ar^{n+1},$$ and so $\sum_{x=1}^n ar^x = (ar - ar^{n+1})/(1-r) = ar \cdot (1-r^n)/(1-r)$.