Q. $\sum_{n = 1}^{n}{\sum_{i = 1}^{i}\Sigma_{j = 1}^{j}}$ is equal to
A) $4\frac{n(n + 1)(2n + 1)}{6}$
B) $\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2}$
C) $\frac{n(n + 1)}{2}$
D) $\frac{n(n + 1)(n + 2)}{6}$
Explanation
(d)
$\sum_{i = 1}^{n}\ \sum_{j = 1}^{i}\ \sum_{k = 1}^{j}{1 = \ }\sum_{i = 1}^{n}\ \sum_{j = 1}^{i}{j\ }$
$= \sum_{i = 1}^{n}\ \frac{i(i + 1)}{2} = \frac{1}{2}\sum_{i = 1}^{n}\ \left( i^{2} + i \right)$
$= \frac{1}{2}\left\lbrack \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right\rbrack$
$= \frac{n(n + 1)}{12}(2n + 4)$
$= \frac{n(n + 1)(n + 2)}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.