<p>Let \(A\) be the sum of the first 20 terms and \(B\) be the sum of the first 40 terms of the series \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \cdots\). If \(B - 2A = 100\lambda\), then \(\lambda\) is equal to</p>
Step-by-Step Solution
Key Concept: Separate the series into two subsequences: odd-positioned terms (1², 3², 5², ...) and even-positioned terms (2·2², 2·4², 2·6², ...), then calculate their sums independently using standard formulas for sum of squares.
<p><strong>Step 1: Identify the pattern</strong></p><p>The series is: 1² + 2·2² + 3² + 2·4² + 5² + 2·6² + ...</p><p>Positions 1,3,5,7,... have coefficient 1: 1², 3², 5², 7²,...</p><p>Positions 2,4,6,8,... have coefficient 2: 2·2², 2·4², 2·6², 2·8²,...</p><p><strong>Step 2: Calculate A (first 20 terms)</strong></p><p>10 odd-positioned terms: 1² + 3² + 5² + ... + 19² = Σ(2k-1)² for k=1 to 10</p><p>Using formula: Σ(2k-1)² = (10)(21)(41)/3 = 2870</p><p>10 even-positioned terms: 2(2² + 4² + 6² + ... + 20²) = 2·4(1² + 2² + 3² + ... + 10²)</p><p>= 8·(10)(11)(21)/6 = 8·385 = 3080</p><p>∴ A = 2870 + 3080 = 5950</p><p><strong>Step 3: Calculate B (first 40 terms)</strong></p><p>20 odd-positioned terms: 1² + 3² + 5² + ... + 39² = Σ(2k-1)² for k=1 to 20</p><p>= (20)(41)(81)/3 = 22140</p><p>20 even-positioned terms: 2(2² + 4² + ... + 40²) = 2·4(1² + 2² + ... + 20²)</p><p>= 8·(20)(21)(41)/6 = 8·2870 = 22960</p><p>∴ B = 22140 + 22960 = 45100</p><p><strong>Step 4: Find λ</strong></p><p>B - 2A = 45100 - 2(5950) = 45100 - 11900 = 33200</p><p>Given: B - 2A = 100λ</p><p>100λ = 33200</p><p>∴ λ = 332</p>
Correct Answer: D