<p>Given series is: \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \cdots\). The sum of first \(n\) terms when \(n\) is even is \(\dfrac{n(n+1)^2}{2}\). Find \(S_n\) for odd \(n\).</p>
<p>\(\dfrac{n^2(n+1)}{2}\)</p>
<p>\(\dfrac{n(n+1)^2}{2}\)</p>
<p>\(\dfrac{n^2(n-1)}{2}\)</p>
<p>\(\dfrac{(n+1)^2(n+2)}{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize the series has alternating pattern: odd-indexed terms are squares, even-indexed terms are 2 times squares. For odd n, express S_n using the given formula for (n+1) and subtract the extra even-indexed term.
<p><strong>Step 1:</strong> Identify the pattern. The series is: $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \cdots$</p><p>Odd positions (1st, 3rd, 5th, ...): $1^2, 3^2, 5^2, \ldots$ (perfect squares)</p><p>Even positions (2nd, 4th, 6th, ...): $2(2^2), 2(4^2), 2(6^2), \ldots$ (2 times even square)</p><p><strong>Step 2:</strong> For even $n$, we are given: $S_n = \dfrac{n(n+1)^2}{2}$</p><p><strong>Step 3:</strong> For odd $n$, write $n = 2k+1$ for some non-negative integer $k$. Then $n+1 = 2k+2$ (even).</p><p>$S_{n+1} = S_{2k+2} = \dfrac{(2k+2)(2k+3)^2}{2} = (k+1)(2k+3)^2$</p><p><strong>Step 4:</strong> The $(n+1)$-th term (which is the $(2k+2)$-th term, an even position) is:</p><p>$a_{n+1} = 2(2k+2)^2 = 2 \cdot 4(k+1)^2 = 8(k+1)^2$</p><p><strong>Step 5:</strong> Therefore: $S_n = S_{n+1} - a_{n+1} = (k+1)(2k+3)^2 - 8(k+1)^2$</p><p>$= (k+1)[(2k+3)^2 - 8(k+1)]$</p><p>$= (k+1)[4k^2 + 12k + 9 - 8k - 8]$</p><p>$= (k+1)(4k^2 + 4k + 1) = (k+1)(2k+1)^2$</p><p><strong>Step 6:</strong> Substitute back $k = \dfrac{n-1}{2}$:</p><p>$S_n = \dfrac{n+1}{2} \cdot n^2 = \dfrac{n^2(n+1)}{2}$</p><p>∴ Answer: A — $S_n = \dfrac{n^2(n+1)}{2}$ for odd $n$</p>
Correct Answer: A