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Question: In \(n = (1999)!\) then \(\sum_{x = 1}^{1999}{\log_{n}x}\) is equal to....

In n=(1999)!n = (1999)! then ∑x=11999log⁡nx\sum_{x = 1}^{1999}{\log_{n}x} is equal to.

A

1

B

0

C

19991999\sqrt[1999]{1999}

D

– 1

Answer

1

Explanation

Solution

12(en+e−n)\frac{1}{2}(e^{n} + e^{- n})

1+1+22!+1+2+33!+1+2+3+44!+....∞=e1 + \frac{1 + 2}{2!} + \frac{1 + 2 + 3}{3!} + \frac{1 + 2 + 3 + 4}{4!} + ....\infty = e.