यदि `n=(2017)!` है, तो `(1)/(log_(2)n)+(1)/(log_(3)n)+(1)/(log_(4)n)+....+(1)/(log_(2017)n)` किसके बराबर है ?

Options

A. 0
B. 1
C. `n/2`
D. `n`

Answer

B

Explanation

`1/(log_(2)n) + 1/(log_(3)n) + 1/(log_(4)n) + .......+1/(log_(2017)n)`
` = log_(n) 2 + log_(n) 3 + log_(n) 4 + ......+log_(n) 2017 [:' log_(b) a = 1/(log_(a)b)]`
` = log_(n) (2 * 3 * 4 * ....2017]`
` = log_(n)(2017)`|
`[:' log_(n) a + log_(n) b = log_(n) ab]`
`[:' n (n-1)(n-2).....3*1*1=n!]`
` = log_((2017)!)(2017)!" "[:' n = (2017)!]`
`= 1 " " [:' log_(a) a = 1]`