<p>The given series is \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \cdots\). If the sum of the first 20 terms is \(A\) and the sum of the first 40 terms is \(B\), then \(B - 2A = 100\lambda\). Find \(\lambda\).</p>
Step-by-Step Solution
Key Concept: Identify the pattern: odd-indexed terms are perfect squares (1², 3², 5², ...) and even-indexed terms are twice perfect squares (2·2², 2·4², 2·6², ...). Separate the series into two subsequences and use summation formulas for sum of squares.
<p><strong>Step 1: Identify the Pattern</strong><br>The series is: 1² + 2·2² + 3² + 2·4² + 5² + 2·6² + ...</p><p>Terms at odd positions (1st, 3rd, 5th, ...): 1², 3², 5², ... = (odd numbers)²<br>Terms at even positions (2nd, 4th, 6th, ...): 2·2², 2·4², 2·6², ... = 2·(even numbers)²</p><p><strong>Step 2: Express Sum of First 20 Terms</strong><br>In 20 terms, there are 10 odd-positioned and 10 even-positioned terms.<br>A = (1² + 3² + 5² + ... + 19²) + 2(2² + 4² + 6² + ... + 20²)</p><p>Sum of first n odd numbers squared: 1² + 3² + 5² + ... + (2n-1)² = n(2n-1)(2n+1)/3<br>For 10 terms: 1² + 3² + ... + 19² = 10(19)(21)/3 = 1330</p><p>Sum of first n even numbers squared: 2² + 4² + 6² + ... + (2n)² = 2n(n+1)(2n+1)/3<br>For 10 terms: 2² + 4² + ... + 20² = 2·10·11·21/3 = 1540</p><p>∴ A = 1330 + 2(1540) = 1330 + 3080 = 4410</p><p><strong>Step 3: Express Sum of First 40 Terms</strong><br>In 40 terms, there are 20 odd-positioned and 20 even-positioned terms.<br>B = (1² + 3² + 5² + ... + 39²) + 2(2² + 4² + 6² + ... + 40²)</p><p>For 20 odd terms: 1² + 3² + ... + 39² = 20(39)(41)/3 = 10660</p><p>For 20 even terms: 2² + 4² + ... + 40² = 2·20·21·41/3 = 11480</p><p>∴ B = 10660 + 2(11480) = 10660 + 22960 = 33620</p><p><strong>Step 4: Calculate B - 2A</strong><br>B - 2A = 33620 - 2(4410) = 33620 - 8820 = 24800</p><p><strong>Step 5: Find λ</strong><br>Given: B - 2A = 100λ<br>24800 = 100λ<br>λ = 248</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B