Question

The sum of the infinite series $1^2-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+\frac{5^2}{5^4}-\frac{6^2}{5^5}+.....$ is

• A. $\frac{1}{2}$
• B. $\frac{25}{24}$
• C. $\frac{25}{54}$
• D. $\frac{125}{252}$

Answer 1

The correct option is C $\frac{25}{54}$

$S=1^2-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....\frac{S}{5}=\frac{1^2}{5}-\frac{2^2}{5^2}-\frac{3^2}{5^3}+...S+\frac{S}{5}=1^2-(\frac{2^2}{5^2}-\frac{1^2}{5})+\left(\frac{3^2-2^2}{5^2}\right)-\left(\frac{4^2-3^2}{5^3}\right)+...\frac{6S}{5}=1-\frac{3}{5}+\frac{5}{5^2}-\frac{7}{5^3}+\frac{9}{5^4}+....\frac{6S}{25}=\frac{1}{5}-\frac{3}{5^2}+\frac{5}{5^3}-\frac{7}{5^4}+....\frac{6S}{5}+\frac{6S}{25}=1-\frac{2}{5}+\frac{2}{5^2}-\frac{2}{5^3}+\frac{2}{5^4}-....\frac{36S}{5}=1-\frac{2}{5}(1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+....)=1-\frac{2}{5}\left(\frac{1}{1+\frac{1}{5}}\right)=1-\frac{2}{5}\left(\frac{5}{6}\right)S=\frac{25}{54}$