Question

The sum of $n$ terms of $1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)$ $+\ldots$ is-

• A. $\frac{ n ( n +1)(2 n +1)}{6}$
• B. $\frac{ n ( n +1)(2 n -1)}{6}$
• C. $\frac{1}{12} n(n+1)^2(n+2)$
• D. $\frac{1}{12} n^2(n+1)^2$

Answer 1

Answer: (C)

Correct answer is (c) $\frac{1}{12} n(n+1)^2(n+2)$