Sequences & Series
Sum of Series with Mixed Terms
Grade 11
Question:
<p>Let <span class="math">A</span> be the sum of the first 20 terms and <span class="math">B</span> be the sum of the first 40 terms of the series <span class="math">1 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \ldots</span></p><p>If <span class="math">B - 2A = 100\lambda</span>, then <span class="math">\lambda</span> is equal to</p>
<p>(A) 496</p>
<p>(B) 232</p>
<p>(C) 248</p>
<p>(D) 464</p>
Step-by-Step Solution
Key Concept: Split the series into odd-indexed and even-indexed terms and use formulas for sum of squares of arithmetic progressions.
<p>The series can be split into two parts: odd-indexed terms \(1^2 + 3^2 + 5^2 + \ldots\) and even-indexed terms \(2 \cdot 2^2 + 2 \cdot 4^2 + 2 \cdot 6^2 + \ldots\)</p><p><strong>For the first 20 terms:</strong> We have 10 odd-indexed and 10 even-indexed terms.</p><p>Sum of first 10 odd squares: \(1^2 + 3^2 + 5^2 + \ldots + 19^2 = \frac{10(2 \cdot 10 - 1)(2 \cdot 10 + 1)}{3} = \frac{10 \cdot 19 \cdot 21}{3} = 1330\)</p><p>Sum of even terms: \(2(2^2 + 4^2 + 6^2 + \ldots + 20^2) = 2 \cdot 4(1^2 + 2^2 + \ldots + 10^2) = 8 \cdot \frac{10 \cdot 11 \cdot 21}{6} = 8 \cdot 385 = 3080\)</p><p>Thus \(A = 1330 + 3080 = 4410\)</p><p><strong>For the first 40 terms:</strong> We have 20 odd-indexed and 20 even-indexed terms.</p><p>Sum of first 20 odd squares: \(1^2 + 3^2 + \ldots + 39^2 = \frac{20(2 \cdot 20 - 1)(2 \cdot 20 + 1)}{3} = \frac{20 \cdot 39 \cdot 41}{3} = 10660\)</p><p>Sum of even terms: \(2(2^2 + 4^2 + \ldots + 40^2) = 8 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 8 \cdot 2870 = 22960\)</p><p>Thus \(B = 10660 + 22960 = 33620\)</p><p>\(B - 2A = 33620 - 2(4410) = 33620 - 8820 = 24800 = 100\lambda\)</p><p>Therefore \(\lambda = 248\)</p><p>∴ Answer is (C).</p>
Correct Answer: C