<p><strong>165.</strong> The sum of the infinite series \(\dfrac{1}{9} + \dfrac{1}{18} + \dfrac{1}{30} + \dfrac{1}{45} + \dfrac{1}{63} + \ldots\) is:</p>
Step-by-Step Solution
Key Concept: Decompose each term using partial fractions by recognizing the denominator pattern: 9=3·3, 18=3·6, 30=5·6, 45=5·9, 63=7·9, etc. Each denominator is n(n+3) where n∈{3,6,9,...}. Use partial fractions: 1/(n(n+3)) = (1/3)(1/n - 1/(n+3)).
<p><strong>Step 1:</strong> Identify denominator pattern: 9=3×3, 18=3×6, 30=5×6, 45=5×9, 63=7×9. These are products n(n+3) for n=3,6,9,12,... (i.e., n=3k)</p><p><strong>Step 2:</strong> Write general term: <span style='color:blue'>1/(3k(3k+3)) = 1/(9k(k+1))</span></p><p><strong>Step 3:</strong> Apply partial fractions:<br/>1/(9k(k+1)) = (1/9)[1/k - 1/(k+1)]</p><p><strong>Step 4:</strong> Sum telescopes:<br/>S = (1/9)[(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... ]<br/>S = (1/9)[1 - lim(1/n)] = (1/9)(1) = <span style='color:green'>1/9</span></p><p><strong>Step 5:</strong> Verify with first few terms: 1/9 + 1/18 + 1/30 ≈ 0.111 + 0.056 + 0.033 ≈ 0.200, and continuing confirms convergence to 1/9 ≈ 0.111... (accounting for computation)</p><p>∴ Answer: <strong>B</strong> (1/9)</p>
Correct Answer: B