Correct Answer - C
We have, `f(x)=sum_(r=1)^(n)(sin""(x)/(2^(r-1))+tan""(x)/(2^(r)))`
`implies f(x)=sum_(r=1)^(n)f_(r)(x), where" " f_(r)(x)= sin""(x)/(2^(r-1))+tan""(x)/(2^(r))`
since each of ` sin""(x)/(2^(r-1)) and (x)/(2^(r ))` are periodic functions with period `2^(r)pi`. Therefore, `f_(r)(x)` is periodic with period ` 2^(r) pi`.
But , `f(x)=sum_(r=1)^(n)f_(r)(x)` .
Therefore, , f(x) is periodic with period T given by
T=LCM of `(2pi, 2^(2)pi, ....,2^(n)pi)=2^(n)pi`