Question:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}\) is equal to
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}\) is equal to
Updated On: Oct 4, 2024
- \(\frac{n}{3n+2}\)
- \(\frac{n}{5n+4}\)
- \(\frac{n}{6n+2}\)
- \(\frac{n}{6n+4}\)
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The Correct Option is D
Solution and Explanation
The correct option is (D): \(\frac{n}{6n+4}\)
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