<p>Find the sum \(3 + 7 + 14 + 24 + 37 + \cdots\) up to 20 terms.</p>
Step-by-Step Solution
Key Concept: Recognize this as a sum of series where differences form an arithmetic progression. The first differences are 4, 7, 10, 13, ... (AP with d=3), so apply the method of differences or find the general term using Tn = an² + bn + c.
<p><strong>Step 1: Find the pattern</strong></p><p>Terms: 3, 7, 14, 24, 37, ...</p><p>First differences: 4, 7, 10, 13, ... (AP with a=4, d=3)</p><p>Second differences: 3, 3, 3, ... (constant)</p><p><strong>Step 2: Find general term Tn</strong></p><p>Since second differences are constant, Tn = an² + bn + c</p><p>T₁ = 3: a + b + c = 3</p><p>T₂ = 7: 4a + 2b + c = 7</p><p>T₃ = 14: 9a + 3b + c = 14</p><p>Solving: 3a + b = 4 and 5a + b = 7</p><p>Therefore: 2a = 3 → a = 3/2, b = 5/2, c = -2</p><p>So Tn = (3/2)n² + (5/2)n - 2 = (3n² + 5n - 4)/2</p><p><strong>Step 3: Find sum of 20 terms</strong></p><p>S₂₀ = Σ(3n²/2 + 5n/2 - 2) from n=1 to 20</p><p>= (3/2)Σn² + (5/2)Σn - 2(20)</p><p>= (3/2)·[20·21·41/6] + (5/2)·[20·21/2] - 40</p><p>= (3/2)·2870 + (5/2)·210 - 40</p><p>= 4305 + 525 - 40</p><p>= 4790 - 40</p><p>= <strong>4240</strong></p>
Correct Answer: 4240