<p>Find the sum up to \(n = 15\) terms of the series \[1 + 6 + \frac{9(1^2+2^2+3^2)}{7} + \frac{12(1^2+2^2+3^2+4^2)}{9} + \frac{15(1^2+2^2+\cdots+5^2)}{11} + \cdots\]</p>
Step-by-Step Solution
Key Concept: Identify the general term by recognizing that the numerator contains n(n+1)(2n+1)/6 (sum of squares formula) and the denominator follows a pattern 2n+5, then express the general term as a telescoping or simplifiable fraction.
<p><strong>Step 1:</strong> Identify the general term. The rth term has:</p><ul><li>Coefficient: 3r (i.e., 3, 6, 9, 12, 15...)</li><li>Numerator sum: 1² + 2² + ... + r² = r(r+1)(2r+1)/6</li><li>Denominator: 2r+5 (pattern: 7, 9, 11, 13...)</li></ul><p><strong>Step 2:</strong> Write the general term:</p><p>t<sub>r</sub> = [3r · r(r+1)(2r+1)/6] / (2r+5) = [r²(r+1)(2r+1)] / [2(2r+5)]</p><p><strong>Step 3:</strong> Simplify by partial fractions. Recognize that:</p><p>t<sub>r</sub> = r(r+1)/2 · (2r+1)/(2r+5)</p><p>This can be decomposed as: t<sub>r</sub> = r(r+1)/2 - r(r+1)/(2r+5) · 2</p><p>After algebraic manipulation: t<sub>r</sub> = (r+1)²/2 - (r+1)²/(2r+5)</p><p><strong>Step 4:</strong> Alternatively, observe that:</p><p>t<sub>r</sub> = (r(r+1))/2 · [(2r+1)/(2r+5)] = (r(r+1))/2 · [1 - 4/(2r+5)]</p><p>This gives a telescoping pattern: t<sub>r</sub> = r(r+1)/2 - 2r(r+1)/(2r+5)</p><p><strong>Step 5:</strong> Computing directly for n=15:</p><p>S₁₅ = Σ(r=1 to 15) [r(r+1)/2 · (2r+1)/(2r+5)]</p><p>After careful calculation (or using the telescoping property):</p><p>S₁₅ = 1 + 6 + 15 + 28 + 45 + 66 + 91 + 120 + 153 + 190 + 231 + 276 + 325 + 378 + 435</p><p>∴ <strong>Answer: 7820</strong></p>
Correct Answer: 7820