<p>Find the sum \(1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + \cdots\) up to the 22nd term.</p>
Step-by-Step Solution
Key Concept: Recognize that the nth term is the sum of squares from 1 to n, which equals n(n+1)(2n+1)/6. The required sum becomes Σ[k(k+1)(2k+1)/6] for k=1 to 22, which can be split into three separate summations of polynomial series.
<p><strong>Step 1:</strong> Recognize the nth term structure. The nth term = 1² + 2² + ... + n² = n(n+1)(2n+1)/6</p><p><strong>Step 2:</strong> Required sum = Σ[n(n+1)(2n+1)/6] for n=1 to 22 = (1/6)Σ[2n³ + 3n² + n]</p><p><strong>Step 3:</strong> Split into parts: = (1/6)[2Σn³ + 3Σn² + Σn]</p><p><strong>Step 4:</strong> Apply formulas for n=1 to 22:</p><p>• Σn = 22(23)/2 = 253</p><p>• Σn² = 22(23)(45)/6 = 3795</p><p>• Σn³ = [22(23)/2]² = 253² = 64009</p><p><strong>Step 5:</strong> Substitute: = (1/6)[2(64009) + 3(3795) + 253]</p><p>= (1/6)[128018 + 11385 + 253]</p><p>= (1/6)[139656]</p><p>= 23276</p><p>∴ Answer: <strong>23276</strong></p>
Correct Answer: 23276