Question

If `(1)/("log"_(2)a) + (1)/("log"_(4)a) + (1)/("log"_(8)a) + (1)/("log"_(16)a) + …. + (1)/("log"_(2^(n))a) = (n(n+1)/(lambda))` then `lambda` equals
A. `"log"_(2)a`
B. `"log"_(a)4`
C. `"log"_(2)a^(2)`
D. none of these

Answer 1

Correct Answer - C
`(1)/("log"_(2)a) + (1)/("log"_(4)a) + (1)/("log"_(8)a) + (1)/("log"_(16)a) + …. + (1)/("log"_(2^(n))a) = (n(n+1)/(lambda))`
`rArr "log"_(a) 2 + "log"_(a) 2^(2) + "log"_(a) 2^(3) + …. +"log"_(a) 2^(n) =(n(n+1)/(lambda)`
`rArr "log"_(a) 2 + 2"log"_(a) 2 + 3"log"_(a) 2 +...+ n"log"_(a) 2 = (n(n+1)/(lambda))`
`rArr (1+2+... +n) "log"_(a)2 = (n(n+1)/(lambda))`
`rArr (n(n+1)/(2))"log"_(a) 2 = (n(n+1)/(lambda))`
`rArr (1)/(2) "log"_(a) 2= (1)/(lambda) rArr lambda = 2 "log"_(2)a = "log"_(2)a^(2)`