Question

The sum \(S_n\) of the first \(n\) terms of the series \[ 1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+\cdots \] is given by

• A. \[ S_n=\frac{n^2(n+1)^2}{2},\; n \text{ is even}; \qquad S_n=\frac{n(n+1)}{2},\; n \text{ is odd} \]
• B. \[ S_n=\frac{n(n^2+1)}{2},\; n \text{ is even}; \qquad S_n=\frac{n^2(n+1)}{2},\; n \text{ is odd} \]
• C. \[ S_n=\frac{n(n+1)^2}{2},\; n \text{ is even}; \qquad S_n=\frac{n^2(n+1)}{2},\; n \text{ is odd} \]
• D. \[ S_n=\frac{n^2(n^2+1)}{2},\; n \text{ is even}; \qquad S_n=\frac{n(n+1)^2}{2},\; n \text{ is odd} \] \bigskip

Hint:

Whenever a series has different patterns for odd and even terms, split the sum into odd-indexed and even-indexed parts. Then apply standard summation formulas separately and simplify.

Answer 1

The Correct Option is C

Concept: Separate the odd and even indexed terms and apply the standard sum of squares formula independently.

Step 1: For even \(n=2m\), \(S_{2m}=\sum_{k=1}^{m}(2k-1)^2+2\sum_{k=1}^{m}(2k)^2\). Using the identities for odd squares and square sums, \(S_{2m}=m(2m+1)^2=\dfrac{n(n+1)^2}{2}\).

Step 2: For odd \(n=2m-1\), subtract the last even term from \(S_{2m}\): \(S_{2m-1}=m(2m+1)^2-8m^2=m(2m-1)(2m)=\dfrac{n^2(n+1)}{2}\).

Step 3: Hence, \(\boxed{S_n=\begin{cases}\dfrac{n(n+1)^2}{2},&n\text{ even}\\[4pt]\dfrac{n^2(n+1)}{2},&n\text{ odd}\end{cases}}\).