Question

The value of the series $\left(\frac{1^1}{2}\right)+\left(\frac{1^2+2^2}{3}\right)+\left(\frac{1^2+2^2+3^2}{4}\right)+.....$ is

• A. $\frac{n(n+1)(4n+5)}{36}$
• B. $\frac{n(n-1)(4n+5)}{6}$
• C. $\frac{n(n+1)(4n+2)}{3}$
• D. None of these

Answer 1

The correct option is A

$\frac{n(n+1)(4n+5)}{36}$

nth term = $\frac{1^2+2^2+3^2+......+n^2}{n+1}\Rightarrow \frac{n(n+1)(2n+1)}{6(n+1)}=\frac{1}{6}(2n^2+n)Sum=\sum nthterm=\frac{1}{6}[2\sum n^2+\sum n]=\frac{1}{6}[\frac{2\left(n\right)(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]=\frac{1}{36}n(n+1)[4n+2+3]=\frac{n(n+1)(4n+5)}{36}$