Question:

<p><strong>Step 1:</strong> Find which group contains the 2000th term. Group n contains n terms, so total terms through group n is: 1+2+3+...+n = n(n+1)/2</p><p>We need n(n+1)/2 ≥ 2000, so n(n+1) ≥ 4000. Testing: 63×64 = 4032 ≥ 4000 ✓ and 62×63 = 3906 < 4000</p><p>So through group 62: 62×63/2 = 1953 terms. The 2000th term is in group 63, at position 2000 - 1953 = 47.</p><p><strong>Step 2:</strong> Sum all complete groups 1 through 62. Group n contributes: n × (1+2+...+n) = n × n(n+1)/2 = n²(n+1)/2</p><p>Sum = Σ(n=1 to 62) n²(n+1)/2 = (1/2)Σ(n=1 to 62)(n³+n²)</p><p><strong>Step 3:</strong> Use formulas: Σn³ = [n(n+1)/2]² and Σn² = n(n+1)(2n+1)/6</p><p>For n=62: Σn³ = (62×63/2)² = 1953² = 3,814,209 and Σn² = 62×63×125/6 = 81,395</p><p>Sum through group 62 = (1/2)(3,814,209 + 81,395) = 1,947,802</p><p><strong>Step 4:</strong> Add contribution from group 63 (47 terms of value 63): 47 × 63 = 2,961</p><p>∴ Total sum = 1,947,802 + 2,961 = <strong>1,950,763</strong></p>