JEE Main · 2004 · 24 Apr 2004 · Offline · Sequences and Series

In this series the odd-numbered terms are $k^2$ and the even-numbered terms are $2k^2$, and we are told the sum to an even number of terms is $\frac{n(n+1{)}^2}{2}$. When $n$ is odd, write the sum to $n$ terms as the sum to $n-1$ terms, an even count, plus the $n$th term. The even formula applied at $n-1$ gives $\frac{(n-1)n^2}{2}$, and the $n$th term sits in an odd position so it equals $n^2$. Adding, $\frac{(n-1)n^2}{2}+n^2=\frac{n^2(n-1)+2n^2}{2}=\frac{n^2(n+1)}{2}$.