Let a_1, a_2, a_3, ... be an infinite geometric series whose sum is 3. Replacing each of the terms of the series by their squares results in a series whose sum is the same. Find the sequence.
+600 Nice.
\(\frac{a}{1-r}=3 \rightarrow a=3-3r\) by geometric sum formula, where r is the common ratio. Our sequence would be \(a,ar,ar^2,ar^3,...\). When squared, it would be \(a^2, a^2r^2, a^2r^4, a^2r^6\). This sequence has first term a^2 and common ratio r^2, so by the geometric sum formula this is \(\frac{a^2}{1-r^2}=3 \rightarrow \frac{(3-3r)^2}{1-r^2}=3\rightarrow 9r^2-12r+9=3-3r^2 \rightarrow r=\frac{1\pm i}{2}\). So also \(a=3-\frac{3\pm 3i}{2}=\frac{3\pm 3i}{2}\).
