Question

The value of the infinite series $\frac{1^2+2^2}{\underset{–}{|3}}+\frac{1^2+2^2+3^2}{\underset{–}{|4}}+\frac{1^2+2^2+3^3+4^2}{\underset{–}{|5}}........$ is

• A. $e$
• B. $5e$
• C. $\frac{5e}{6}-\frac{1}{2}$
• D. $\frac{5e}{6}$

Answer 1

The correct option is C $\frac{5e}{6}-\frac{1}{2}$

$T_n=\frac{n(n+1)(2n+1)}{6(n+1)!}S={\sum }_{i=2}^{\infty }\frac{n(n+1)(2n+1)}{6(n+1)!}={\sum }_{i=2}^{\infty }\frac{(2n+1)}{6(n-1)!}S={\sum }_{i=2}^{\infty }\frac{2n-2+3}{6(n-1)!}S=\frac{1}{6}{\sum }_{i=2}^{\infty }\frac{2(n-1)}{(n-1)!}+\frac{3}{6}{\sum }_{i=2}^{\infty }\frac{1}{(n-1)!}S=\frac{2}{6}{\sum }_{i=2}^{\infty }\frac{1}{(n-2)!}+\frac{1}{2}{\sum }_{i=2}^{\infty }\frac{1}{(n-1)!}S=\frac{1}{3}e+\frac{1}{2}(e-1)S=\frac{5e}{6}-\frac{1}{2}$