Find the natural number `a` for which `sum_(k=1)^nf(a+k)=16(2^n-1),` where the function `f` satisfies the relation `f(x+y)=f(x)f(y)` for all natural number `x , ya n d ,fu r t h e r ,f(1)=2.`

Options

A. 16
B. 3
C. 4
D. 2

Answer

B

Explanation

`f(x+y)=f(x)*f(y) rArr f(x)=a^(x)`
`f(1)=2 rArr f(x)=2^(x)`
`underset(k=1)overset(10)sum f(a+k)=2^(a)[2+2^(2)+...+2^(10)]=(2^(a+1)(z^(10)-1))/(2-1)`
Also, `2^(a+1)(2^(10)-1)=16(2^(10)-1)`
`rArr a+1=4 rArr a=3`