Question

If `n=(2017)!`, then what is
`(1)/(log_(2)n)+(1)/(log_(3)n)+(1)/(log_(4)n)+....+(1)/(log_(2017)n)` equal to?

Answer 1

Correct Answer - B
`n=(2017)!`
`(1)/(log_(2)n)+(1)/(log_(3)n)+(1)/(log_(4)n)+....+(1)/(log_(2017)n)`
`=(1)/((log_(n))/(log_(2)))+(1)/((log_(n))/(log_(3)))+(1)/((log_(n))/(log_(4)))+......+(1)/(log_(n)/(log_(2017)))(becauselog_(a)b=(log_(b))/(log_(a)))`
`=(log_(2))/(log_(n))+(log_(3))/(log_(n))+(log_(4))/(log_(n))+......+(log_(2017))/(log_(n))`
`=(log_(2)+log_(3)+log_(4)+......+log_(2017))/(log_(n))`
`(log(2.3.4....2017))/(log_(n))" "(because loga+logb+logc+.....=loga.b.c.....)`
`=(log(2017!))/(log_(n))=(log_(n))/(log_(n))=1`