Question:medium
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:
Show Hint
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
- \( \frac{1}{17} \)
- \( \frac{1}{18} \)
- \( \frac{1}{19} \)
- \( \frac{1}{20} \)
- \( \frac{1}{21} \)
Show Solution
The Correct Option is D
Solution and Explanation
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