Question

If $\frac{25}{k}=1^2-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+\frac{5^2}{5^4}-\frac{6^2}{5^2}+......\infty$, then find the value of $k$.

Answer 1

$S=1^2-\frac{2^2}{5}+\frac{3^3}{5^2}-\frac{4^2}{5^3}+....\frac{S}{5}=\frac{1^2}{5}-\frac{2^2}{5^2}+\frac{3^2}{5^2}+....\frac{65}{5}=1^2-(\frac{2^2}{5}-\frac{1^2}{5})+\left(\frac{3^2-2^2}{5^2}\right)-\left(\frac{4^2-3^2}{5^3}\right)....\frac{65}{5}=1-\frac{3}{5}-\frac{5}{5^2}-\frac{7}{5^3}+\frac{9}{5^4}-.....\frac{65}{25}=\frac{1}{5}-\frac{3}{5^2}-\frac{5}{5^3}+\frac{7}{5^4}+....\frac{65}{5}+\frac{65}{25}=1-\frac{2}{5}-\frac{2}{5^2}-\frac{2}{5^3}+\frac{2}{5^4}+...\frac{365}{25}=1-\frac{2}{5}(1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+....)=1-\frac{2}{5}\left(\frac{1}{1+\frac{1}{5}}\right)\frac{365}{25}=\frac{2}{3}S=\frac{25}{54}=\frac{25}{k}k=54$