Question

The sum of the series
$S=1+\frac{4}{5}+\frac{9}{25}+\frac{16}{125}+\cdots \infty$
is $\frac{15m}{32}$, then the value of $m$ is

Answer 1

$\text{Let}a=\frac{1}{5}\text{then}S=1+4a+9a^2+16a^3+\cdots \infty \cdots \left(1\right)aS=a+4a^2+9a^3+16a^4+\cdots \infty \cdots \left(2\right)\left(1\right)-\left(2\right)(1-a)S=1+3a+5a^2+7a^3+\cdots \infty \cdots \left(3\right)(1-a)aS=a+3a^2+5a^3+7a^4+\cdots \infty \cdots \left(4\right)\left(3\right)-\left(4\right)(1-a{)}^{2}S=1+2a+2a^2+2a^3+\cdots \infty \Rightarrow (1-a{)}^{2}S=1+\frac{2a}{1-a}\Rightarrow S=\frac{1+a}{(1-a{)}^3}=\frac{1+\frac{1}{5}}{{(1-\frac{1}{5})}^3}=\frac{75}{32}\therefore m=5$