<p><strong>165.</strong> The sum of the infinite series \(\dfrac{1}{9} + \dfrac{1}{18} + \dfrac{1}{30} + \dfrac{1}{45} + \dfrac{1}{63} + \ldots\ldots\)</p>
<p>(a) \(\dfrac{1}{3}\)</p>
<p>(b) \(\dfrac{1}{4}\)</p>
<p>(c) \(\dfrac{1}{5}\)</p>
<p>(d) \(\dfrac{2}{3}\)</p>
Step-by-Step Solution
Key Concept: Recognize that each denominator follows the pattern n(n+2) for n=3,4,5,... and use partial fraction decomposition: 1/[n(n+2)] = 1/2[1/n - 1/(n+2)]
<p><strong>Step 1: Identify the pattern in denominators</strong></p><p>9 = 3×3, 18 = 3×6, 30 = 5×6, 45 = 5×9, 63 = 7×9</p><p>Rewrite: 9 = 3(3+0), 18 = 3(3+3), 30 = 5(5+1)... Actually: 9 = 3·3, 18 = 3·6 = 3·(3+3), 30 = 5·6, 45 = 5·9, 63 = 7·9</p><p><strong>Better approach:</strong> 1/9 = 1/(3·3), 1/18 = 1/(3·6), 1/30 = 1/(5·6), 1/45 = 1/(5·9), 1/63 = 1/(7·9)</p><p>General term: 1/[n(n+2)] where n = 3,5,7,9,... (odd numbers ≥3)</p><p><strong>Step 2: Apply partial fractions</strong></p><p>1/[n(n+2)] = 1/2[1/n - 1/(n+2)]</p><p><strong>Step 3: Write the telescoping series</strong></p><p>S = 1/2[(1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + (1/9 - 1/11) + ...]</p><p><strong>Step 4: Sum the telescoping series</strong></p><p>S = 1/2 · lim[1/3 - 1/n] = 1/2 · 1/3 = 1/6</p><p>∴ Answer: B (which equals 1/6)</p>
Correct Answer: B