2. The sum of the terms of an infinite geometric series is 2 and the sum of the squares of the terms of this series is 6 . Find the sum of the cubes of the terms of this series.
Answer: $\frac{96}{7}$
Solution: Let $a$ be the first term and let $r \in(-1,1)$ be the common ratio of such infinite geometric series. Then, $a /(1-r) \stackrel{(1)}{=} 2$ and $a^{2} /\left(1-r^{2}\right) \stackrel{(2)}{=} 6$. Squaring (1) gives $a^{2} /(1-r)^{2}=4$ and using (2) yields $(1-r) /(1+r)=3 / 2$. Solving for $r$, we get $r=-\frac{1}{5}$ so from (1) we get $a=2(1-r)=\frac{12}{5}$. Thus,
$$
\frac{a^{3}}{1-r^{3}}=\frac{\frac{12^{3}}{5^{3}}}{1+\frac{1}{5^{3}}}=\frac{12^{3}}{5^{3}+1}=\frac{12 \cdot 12^{2}}{6(25-5+1)}=\frac{288}{21}=\frac{96}{7} .
$$