Question

If $S=1+\frac{8}{3}+\frac{27}{9}+\frac{64}{27}+\cdots \infty ,$ then the value of $8S$ is

Answer 1

Let $a=\frac{1}{3}$
Then, $S=1+8a+27a^2+64a^3+\cdots \infty \cdots \left(1\right)aS=a+8a^2+27a^3+64a^4+\cdots \infty \cdots \left(2\right)\left(1\right)-\left(2\right)(1-a)S=1+7a+19a^2+37a^3+\cdots \infty \cdots \left(3\right)(1-a)aS=a+7a^2+19a^3+37a^4+\cdots \infty \cdots \left(4\right)$

$\left(3\right)-\left(4\right)(1-a{)}^{2}S=1+6a+12a^2+18a^3+24a^4+\cdots \infty \cdots (5\left)\right(1-a{)}^{2}aS=a+6a^2+12a^3+18a^4+\cdots \infty \cdots \left(6\right)$

$\left(5\right)-\left(6\right)(1-a{)}^{3}S=1+5a+6a^2+6a^3+6a^4+\cdots \infty \Rightarrow (1-a{)}^{3}S=1+5a+\frac{6a^2}{1-a}$
$\Rightarrow S=\frac{1+4a+a^2}{(1-a{)}^4}\Rightarrow S=\frac{1+\frac{4}{3}+{\left(\frac{1}{3}\right)}^2}{{(1-\frac{1}{3})}^4}\Rightarrow S=\frac{99}{8}\Rightarrow 8S=99$