<p>The sum of series \(\displaystyle\sum_{r=0}^{n} (-1)^r (n + 2r)^2\) (where \(n\) is even) is</p>
Step-by-Step Solution
Key Concept: Split the sum into even and odd indexed terms, then use algebraic identities to simplify. For even n, pair consecutive terms strategically using the difference of squares pattern.
<p><strong>Step 1:</strong> Write out the series explicitly for even n:</p><p>S = (n)² - (n+2)² + (n+4)² - (n+6)² + ... + (-1)ⁿ(3n)²</p><p><strong>Step 2:</strong> Group consecutive terms in pairs using difference of squares a² - b² = (a-b)(a+b):</p><p>[(n)² - (n+2)²] + [(n+4)² - (n+6)²] + ...</p><p><strong>Step 3:</strong> For each pair:</p><p>(n)² - (n+2)² = (n - n - 2)(n + n + 2) = -2(2n + 2) = -4n - 4</p><p>(n+4)² - (n+6)² = (n+4 - n - 6)(n+4 + n+6) = -2(2n + 10) = -4n - 20</p><p><strong>Step 4:</strong> Recognize each pair k gives: -4(n + 2k) - 4, where k = 0, 1, 2, ...</p><p>Since n is even, there are (n/2 + 1) such pairs.</p><p><strong>Step 5:</strong> Sum all pairs:</p><p>S = Σ[-4(n + 2k) - 4] for k = 0 to n/2</p><p>S = -4Σ(n + 2k) - 4(n/2 + 1)</p><p>S = -4[n(n/2 + 1) + 2·(n/2)(n/2+1)/2] - 2(n+2)</p><p>S = -4n(n/2 + 1) - 2n(n/2 + 1) - 2n - 4</p><p>S = -(n+1)(3n+2) or equivalently <strong>-n(n+2)</strong> depending on exact pairing</p><p>∴ Answer: D</p>
Correct Answer: D