Question

What is `(1)/(log_(2)N)+(1)/(log_(3)N)+(1)/(log_(4)N)+....+(1)/(log_(100)N) " equal to "(Nne1)`?
A. `(1)/(log_(100!)N)`
B. `(1)/(log_(99!)N)`
C. `(99)/(log_(100!)N)`
D. `(99)/(log_(99!)N)`

Answer 1

Correct Answer - A
`(1)/(log_(2)N)+(1)/(log_(3)N)+(1)/(log_(4)N)+....+(1)/(log_(100)N)`
`(1)/((logN)/(log_(2)))+(1)/((logN)/(log_(3)))+(1)/((logN)/(log_(4)))+......+(1)/((logN)/(log_(100)))" "(because log_(a)b=(logb)/(loga))`
`=(log_(2))/(logN)+(log_(3))/(logN)+(log_(4))/(logN)+......+(log_(100))/(logN)`
`=(log_(2)+log_(3)+log_(4)+....+log_(100))/(logN)`
`=(log(2.3.4....100))/(logN)`
`=(log100!)/(logN)=(1)/((logN)/(log100!))=(1)/(log_(100!)N)`