`1/(log_2 a)+1/(log_4 a)+1/(log_8 a)+...` up to `n` terms = `(n(n+1))/k` then `k=`

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))/(k) then k log_(a)2 is equal to

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))/(k) then k log_(a)2 is equal to

The sum of the series: 1/((log)_2 4)+1/((log)_4 4)+1/((log)_8 4)++1/((log)_(2n)4) is (n(n+1))/2 (b) (n(n+1)(2n+1))/(12) (c) (n(n+1))/4 (d) none of these

The sum of the series: (1)/(log_(2)4)+(1)/(log_(4)4)+(1)/(log_(8)4)+...+(1)/(log_(2n)4) is (n(n+1))/(2) (b) (n(n+1)(2n+1))/(12) (c) (n(n+1))/(4) (d) none of these

Prove log a + log a^(2) + "log" a^(3) +……..+ log a^(n) = (n (n + 1))/(2) log a

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

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

Series (1)/(log_(2)^(2)) + (1)/(log_(4)^(4)) + (1)/(log_(8)^(4)) + …..+ (1)/(log_(2^(n))4) =……….