Sum of Geometric Progression

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[edit] Theorem

Let x \in \R, x \ne 1 and n \in \N^*.

Then \sum_{j = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}.


[edit] Corollary

Let a, ar, ar^2, \ldots, ar^{n-1} be a geometric progression.

Then \sum_{j = 0}^{n - 1} ar^j = \frac {a \left({r^n - 1}\right)} {r - 1}.


[edit] Proof

Let S_n = \sum_{j = 0}^{n - 1} x^j = 1 + x + x^2 + \cdots + x^{n-1}.

Then x S_n = x \sum_{j = 0}^{n - 1} x^j = x + x^2 + x^3 + \cdots + x^{n-1} + x^n.

Then x S_n - S_n = -1 + x - x + x^2 - x^2 + \cdots + x^{n-1} - x^{n-1} + x^n = x^n - 1.


The result follows.

[edit] Using Sum Notation

S_n(x - 1) = x S_n - S_n = x \sum_{j = 0}^{n - 1} x^j - \sum_{j = 0}^{n - 1} x^j = \sum_{j = 1}^{n} x^j - \sum_{j = 0}^{n - 1} x^j = x^n + \sum_{j = 1}^{n-1} x^j - \left({x^0 + \sum_{j = 1}^{n - 1} x^j}\right) = x^n - x^0 = x^n - 1

The result follows.


[edit] Alternative Proof

You can directly use the result Difference of Two Powers by setting a = 1 and b = x.


[edit] Proof of Corollary

Follows immediately from the fact that a + ar + ar^2 + \cdots + ar^{n-1} is exactly the same as a \left({1 + r + r^2 + \cdots + r^{n-1}}\right).


[edit] Comment

Note that when x < 1 the result is usually given as \sum_{j = 0}^{n - 1} x^j = \frac {1 - x^n} {1 - x}.


[edit] Sources

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