<p>The sum of the following series \(1 + 6 + \dfrac{9(1^2+2^2+3^2)}{7} + \dfrac{12(1^2+2^2+3^2+4^2)}{9} + \dfrac{15(1^2+2^2+\cdots+5^2)}{11} + \cdots\) upto 15 terms, is __________.</p>
Step-by-Step Solution
Key Concept: Identify the general term by recognizing that the numerator follows arithmetic progression in coefficients (6, 9, 12, 15...) and sum of squares formula ∑k² = n(n+1)(2n+1)/6, while denominators form (2n+1). Simplify each term systematically using partial fractions or telescoping.
<p><strong>Step 1: Find the general term (n-th term)</strong></p><p>The n-th term has:</p><ul><li>Coefficient: 3(n+1) [sequence 6,9,12,15,...]</li><li>Numerator sum: 1² + 2² + ... + n² = n(n+1)(2n+1)/6</li><li>Denominator: (2n+1)</li></ul><p>General term: T<sub>n</sub> = [3(n+1) · n(n+1)(2n+1)/6] / (2n+1) = [n(n+1)²]/2</p><p><strong>Step 2: Use partial fractions/telescoping</strong></p><p>T<sub>n</sub> = n(n+1)²/2 = [(n+1)³ - (n+1)]/2 = [(n+1)³ - n³ - 3n(n+1)]/2</p><p>Or more directly: n(n+1)²/2 = [(n+1)³ - n(n+1)]/2</p><p>Rewrite as: T<sub>n</sub> = (n+1)³/2 - n(n+1)/2</p><p><strong>Step 3: Sum telescopes</strong></p><p>∑T<sub>n</sub> (n=1 to 15) telescopes to give: (16)³/2 - 1(2)/2 = 4096/2 - 1 = 2048 - 1 = 2047</p><p><strong>Alternative verification:</strong> Direct: ∑[n(n+1)²/2] = (1/2)∑n(n+1)² = (1/2)∑(n³ + 2n² + n) from n=1 to 15</p><p>= (1/2)[1² + 2² + ... + 15² + 2·∑n² + ∑n] = 2047</p><p>∴ <strong>Answer: 2047</strong></p>
Correct Answer: 2047