Sequences & Series
Sum of Series
Grade 11
Question:
<p>The sum of the first \(n\) terms of the series \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \ldots\) when \(n\) is even is given. When \(n\) is odd, the sum is</p>
<p>(A) \(\frac{n(n+1)^2}{4}\)</p>
<p>(B) \(\frac{n^2(n+1)}{4}\)</p>
<p>(C) \(\frac{n(n+1)}{2}\)</p>
<p>(D) \(\frac{n^2(n+2)}{4}\)</p>
Step-by-Step Solution
Key Concept: Separate the series into odd and even indexed terms and use formulas for sum of squares.
<p><strong>Solution:</strong> Split the series into odd and even indexed terms.</p><p>Odd terms: \(1^2 + 3^2 + 5^2 + \ldots\) (sum of squares of odd numbers)</p><p>Even terms: \(2 \cdot 2^2 + 2 \cdot 4^2 + 2 \cdot 6^2 + \ldots = 2(2^2 + 4^2 + 6^2 + \ldots)\)</p><p>For odd \(n = 2k-1\), we have \(k\) odd terms and \(k-1\) even terms.</p><p>Sum of first \(k\) odd squares: \(\frac{k(2k-1)(2k+1)}{3}\)</p><p>Sum becomes \(\frac{n(n+1)^2}{4}\) for odd \(n\)</p><p>∴ Answer is A.</p>
Correct Answer: A