Question:
The sum of the series \( \sum_{n=8}^{17} \frac{1}{(n+2)(n+3) \) is equal to:}
The sum of the series \( \sum_{n=8}^{17} \frac{1}{(n+2)(n+3) \) is equal to:}
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In a simple telescoping sum like $\sum_{n=a}^{b} [f(n) - f(n+1)]$, the result is always $f(a) - f(b+1)$.
Updated On: May 6, 2026
- \( \frac{1}{17} \)
- \( \frac{1}{18} \)
- \( \frac{1}{19} \)
- \( \frac{1}{20} \)
- \( \frac{1}{21} \)
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The Correct Option is D
Solution and Explanation
Concept:
This is a telescoping series. We use partial fraction decomposition to simplify the general term:
\[ \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3} \]
Step 1: Write out the terms of the sum.
\[ S = \left( \frac{1}{10} - \frac{1}{11} \right) + \left( \frac{1}{11} - \frac{1}{12} \right) + \dots + \left( \frac{1}{19} - \frac{1}{20} \right) \]
Step 2: Cancel common terms.
The series telescopes, leaving only the first and last parts: \[ S = \frac{1}{10} - \frac{1}{20} \]
Step 3: Final calculation.
\[ S = \frac{2 - 1}{20} = \frac{1}{20} \]
Step 1: Write out the terms of the sum.
\[ S = \left( \frac{1}{10} - \frac{1}{11} \right) + \left( \frac{1}{11} - \frac{1}{12} \right) + \dots + \left( \frac{1}{19} - \frac{1}{20} \right) \]
Step 2: Cancel common terms.
The series telescopes, leaving only the first and last parts: \[ S = \frac{1}{10} - \frac{1}{20} \]
Step 3: Final calculation.
\[ S = \frac{2 - 1}{20} = \frac{1}{20} \]
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