Sequences & Series
Sum of Series
Grade 11
Question:
<p>Let \(A\) be the sum of the first 20 terms and \(B\) be the sum of the first 40 terms of the series \(1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + \ldots\) If \(B - 2A = 100\lambda\), then \(\lambda\) is equal to</p>
<p>496</p>
<p>232</p>
<p>248</p>
<p>464</p>
Step-by-Step Solution
Key Concept: Recognize the series has alternating pattern: odd squares with coefficient 1 and even squares with coefficient 2. Separate into two subsequences and use formulas for sum of squares to find A and B independently.
<p><strong>Step 1: Identify the pattern</strong></p><p>The series is: 1² + 2(2²) + 3² + 2(4²) + 5² + 2(6²) + ...</p><p>Pattern: odd positions have odd squares (coefficient 1), even positions have even squares (coefficient 2).</p><p><strong>Step 2: Write first 20 terms (A)</strong></p><p>In first 20 terms: 10 odd-indexed terms and 10 even-indexed terms</p><p>A = (1² + 3² + 5² + ... + 19²) + 2(2² + 4² + 6² + ... + 20²)</p><p><strong>Step 3: Calculate sum of odd squares</strong></p><p>Sum of first n odd squares: 1² + 3² + ... + (2n-1)² = n(2n-1)(2n+1)/3</p><p>For n=10: (10)(19)(21)/3 = 1330</p><p><strong>Step 4: Calculate sum of even squares</strong></p><p>2² + 4² + ... + (2n)² = 4(1² + 2² + ... + n²) = 4·n(n+1)(2n+1)/6</p><p>For n=10: 4·(10)(11)(21)/6 = 1540</p><p>Therefore: A = 1330 + 2(1540) = 1330 + 3080 = 4410</p><p><strong>Step 5: Calculate sum of first 40 terms (B)</strong></p><p>In first 40 terms: 20 odd-indexed terms and 20 even-indexed terms</p><p>Odd squares (1² + 3² + ... + 39²) with n=20: (20)(39)(41)/3 = 10660</p><p>Even squares (2² + 4² + ... + 40²) with n=20: 4·(20)(21)(41)/6 = 11480</p><p>Therefore: B = 10660 + 2(11480) = 10660 + 22960 = 33620</p><p><strong>Step 6: Find λ</strong></p><p>B - 2A = 33620 - 2(4410) = 33620 - 8820 = 24800</p><p>Given: B - 2A = 100λ</p><p>100λ = 24800</p><p>∴ λ = 248</p>
Correct Answer: C