Question

The sum to n terms of the series $1^2+(1^2+3^2)+(1^2+3^2+5^2)+\cdots$ is:

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

Answer 1

The correct option is C $\frac{1}{6}n(n+1)(2n^2+2n-1)$

Put n = 2 and 3 and then check for the correct choice
Sum of 2 terms = 11
and Sum of 3 terms = 46
at n = 2, $\frac{1}{6}\times 2\left(3\right)\left(11\right)=11$
and at n = 3, $\frac{1}{6}\times 3\times 4\times 23=46$
Hence, choice (c) is correct.