+0

# geometric progression

0
80
1

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.

Sep 12, 2020

#1
+141
+1

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 $$a=3-\frac{3\pm 3i}{2}=\frac{3\pm 3i}{2}$$.

Sep 12, 2020