Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.2 Junior Find the answer

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 - r}$ where $a$ denotes the first term and $- 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:
$\textbf{(A)}\ \frac {a^2}{(1 - r)^2}$ $\textbf{(B)}\ \frac {a^2}{1 + r^2}$ $\textbf{(C)}\ \frac {a^2}{1 - r^2}$ $\textbf{(D)}\ \frac {4a^2}{1 + r^2}$ $\textbf{(E)}\ \text{none of these}$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

Let the original geometric series be $a,ar,ar^2,ar^3,ar^4\cdots$ . Therefore, their squares are $a^2,a^2r^2,a^2r^4,a^2r^6,\cdots$ , which is a geometric sequence with first term $a^2$ and common ratio $r^2$ . Thus, the sum is $\boxed{\textbf{(C)}\ \frac {a^2}{1 - r^2}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.