Question:
The value of \(\frac{1}{log_2N}+\frac{1}{log_3N}+\frac{1}{log_4N}+....\frac{1}{log_15N}\) is
The value of \(\frac{1}{log_2N}+\frac{1}{log_3N}+\frac{1}{log_4N}+....\frac{1}{log_15N}\) is
Updated On: Apr 27, 2024
- \(\frac{1}{log_15^2N}\)
- \(log_N(15!)^2\)
- \(\frac{1}{log_N}(15!)\)
- \(log_N(15)^2\)
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The Correct Option is A
Solution and Explanation
The correct option is(A): \(\frac{1}{log_15^2N}\)
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