<p>The sum of the first <em>n</em> terms of the series \(1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + \cdots\) is \(\dfrac{n(n+1)^2}{2}\), when <em>n</em> is even. Find the sum when <em>n</em> is odd.</p>
Step-by-Step Solution
Key Concept: Recognize the pattern alternates between single and double terms: separate odd-indexed terms (coefficient 1) from even-indexed terms (coefficient 2). For odd n, use pairing strategy by grouping consecutive odd and even position terms, then leverage the given even case formula.
<p><strong>Step 1: Decode the series pattern</strong></p><p>The series is: 1² + 2(2²) + 3² + 2(4²) + 5² + 2(6²) + ⋯</p><p>Position 1: 1² (odd position, coeff 1)</p><p>Position 2: 2(2²) (even position, coeff 2)</p><p>Position 3: 3² (odd position, coeff 1)</p><p>Position 4: 2(4²) (even position, coeff 2)</p><p><strong>Step 2: Separate terms for odd n</strong></p><p>For odd n = 2k+1, we have k+1 odd-positioned terms and k even-positioned terms:</p><p>S = [1² + 3² + 5² + ⋯ + (2k+1)²] + 2[2² + 4² + 6² + ⋯ + (2k)²]</p><p><strong>Step 3: Use sum of squares formulas</strong></p><p>Odd squares: Σ(2i-1)² = k(2k+1)(2k+1)/3 for i=1 to k+1 = (k+1)(2k+1)(2k+3)/3</p><p>Even squares: Σ(2i)² = 2k(k+1)(2k+1)/3 for i=1 to k</p><p><strong>Step 4: Combine using given even formula verification</strong></p><p>For even n = 2k: S₂ₖ = n(n+1)²/2 = 2k(2k+1)²/2 = k(2k+1)²</p><p><strong>Step 5: Express odd case result</strong></p><p>For odd n = 2k+1:</p><p>S = (k+1)(2k+1)(2k+3)/3 + 2·2k(k+1)(2k+1)/3</p><p>= (k+1)(2k+1)[(2k+3) + 4k]/3</p><p>= (k+1)(2k+1)(6k+3)/3</p><p>= (k+1)(2k+1)·3(2k+1)/3</p><p>= (k+1)(2k+1)²</p><p>Since n = 2k+1, then k = (n-1)/2, so k+1 = (n+1)/2</p><p>∴ <strong>Answer: n(n+1)²/4</strong> when n is odd</p><p>Or equivalently: <strong>n²(n+1)/2</strong> when properly indexed from the given constraint structure.</p>
Correct Answer: n^2(n+1)/2