<p>If \(\displaystyle\sum_{r=1}^{n} T_r = \frac{n}{8}(n+1)(n+2)(n+3)\), then find \(\displaystyle\sum_{r=1}^{n}\frac{1}{T_r}\).</p>
Step-by-Step Solution
Key Concept: Find the general term T_r by computing S_n - S_(n-1), then decompose 1/T_r using partial fractions to create a telescoping series.
<p><strong>Step 1: Find General Term T_r</strong></p><p>Given: ∑T_r = n/8(n+1)(n+2)(n+3)</p><p>For n ≥ 2: T_n = S_n - S_(n-1) = n(n+1)(n+2)(n+3)/8 - (n-1)n(n+1)(n+2)/8</p><p>T_n = [n(n+1)(n+2)/8][(n+3) - (n-1)] = n(n+1)(n+2)·4/8 = n(n+1)(n+2)/2</p><p>Check n=1: T_1 = 1·2·3/2 = 3, and S_1 = 1·2·3·4/8 = 3 ✓</p><p><strong>Step 2: Decompose 1/T_r Using Partial Fractions</strong></p><p>1/T_r = 2/[r(r+1)(r+2)]</p><p>Using partial fractions: 2/[r(r+1)(r+2)] = 1/[r(r+1)] - 1/[(r+1)(r+2)]</p><p><strong>Step 3: Sum as Telescoping Series</strong></p><p>∑(r=1 to n) 1/T_r = ∑(r=1 to n) {1/[r(r+1)] - 1/[(r+1)(r+2)]}</p><p>This telescopes:</p><p>= 1/(1·2) - 1/(2·3) + 1/(2·3) - 1/(3·4) + ... + 1/[n(n+1)] - 1/[(n+1)(n+2)]</p><p>= 1/2 - 1/[(n+1)(n+2)]</p><p>= [(n+1)(n+2) - 2]/[2(n+1)(n+2)]</p><p>= [n² + 3n + 2 - 2]/[2(n+1)(n+2)]</p><p>= n(n+3)/[2(n+1)(n+2)]</p><p><strong>∴ Answer: n(n+3)/[2(n+1)(n+2)]</strong></p>
Correct Answer: n(n+3)/(2(n+1)(n+2))