sequences and series - Geometric progression — Sum of terms, sum of terms' cubes, sum of terms' squares

Consider the infinite geometric progression $a_1, a_2, a_3, \dots$. Given the sum of its terms, as well as the sum of the terms' cubes

$a_1 + a_2 + a_3 + \cdots = 3\\ a_1^3 + a_2^3 + a_3^3 + \cdots = \frac{108}{13}$

find the sum of the terms' squares

$a_1^2 + a_2^2 + a_3^2 + \cdots$

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This is what I have so far:

$a_1 + a_2 + a_3 + \cdots = 3\\ \Longrightarrow 1 + q + q^2 + \cdots = \frac{3}{a_1}$

$a_1^3 + a_2^3 + a_3^3 + \cdots = \frac{108}{13}\\ \Longrightarrow 1 + q^3 + q^6 + \cdots = \frac{108}{13a_1}$

$a_1^2 + a_2^2 + a_3^2 + \cdots\\ = \frac{a_1^3}{a_1} + \frac{a_2^3}{a_2} + \frac{a_3^3}{a_3} + \cdots\\ = \frac{a_1^3}{a_1} \left(1 + \frac{q^3}{q} + \frac{q^6}{q^2} + \cdots\right)$

Any help will be much appreciated.

Answer

HINT:

Let the first term be $a$ and the common ratio be $r$

For convergence we need $|r|<1$

So, the sum of the first series $\sum_{0\le n<\infty}a\cdot r^n=\frac a{1-r}=3\ \ \ \ (1)$

So, the sum of the second series $\sum_{0\le n<\infty}a^3\cdot (r^3)^n=\frac {a^3}{1-r^3}=\frac{108}{13}\ \ \ \ (2)$

Cube the first & divide by $(2)$ to get $r$ and then get $a$ from $(1)$

We need $$\sum_{0\le n<\infty}a^2\cdot (r^2)^n=\frac{a^2}{1-r^2}$$