Sequences & Series
Sum of Series
Grade 11
Question:
<p>321. If \(\displaystyle\sum_{k=1}^{n}\left(\sum_{m=1}^{k} m^2\right) = an^4 + bn^3 + cn^2 + dn + e\), then</p>
<p>(a) \(a = \frac{1}{12}\)</p>
<p>(b) \(b = \frac{1}{6}\)</p>
<p>(c) \(d = \frac{1}{6}\)</p>
<p>(d) \(e = 0\)</p>
Step-by-Step Solution
Key Concept: Use the formula for sum of squares ∑(m=1 to k) m² = k(k+1)(2k+1)/6, then sum this expression from k=1 to n to find coefficients of the resulting polynomial.
<p><strong>Step 1:</strong> Use the standard formula: ∑<sub>m=1</sub><sup>k</sup> m² = k(k+1)(2k+1)/6</p><p><strong>Step 2:</strong> Expand this: k(k+1)(2k+1)/6 = (2k³ + 3k² + k)/6</p><p><strong>Step 3:</strong> Now compute ∑<sub>k=1</sub><sup>n</sup> [(2k³ + 3k² + k)/6]</p><p><strong>Step 4:</strong> Separate the sum: (1/6)[2∑k³ + 3∑k² + ∑k]</p><p><strong>Step 5:</strong> Apply standard formulas:</p><ul><li>∑k = n(n+1)/2</li><li>∑k² = n(n+1)(2n+1)/6</li><li>∑k³ = [n(n+1)/2]²</li></ul><p><strong>Step 6:</strong> Substitute and expand:</p><p>(1/6)[2·n²(n+1)²/4 + 3·n(n+1)(2n+1)/6 + n(n+1)/2]</p><p><strong>Step 7:</strong> Simplify to get: (1/12)n(n+1)(n+2)(3n+5)</p><p><strong>Step 8:</strong> Expand: (1/12)(3n⁴ + 12n³ + 19n² + 12n)</p><p>= (1/4)n⁴ + n³ + (19/12)n² + n</p><p>∴ Answer: D</p>
Correct Answer: D