The natural number `a` for which `sum_(k=1)^n f(a+k)=16(2^n-1)` where the function f satisfies the relation `f(x+y)=f(x).f(y)` for all natural numbers x,y and further `f(1)=2` is:-
A) 2 B) 3 C) 1 D) none of these

Explanation

Given, `f(x+y)=f(x)f(y)" " "....(i)"`
and `f(1)=2" " "……(ii)"`
On putting `x=y=1` in Eq. (i), we get
`f(1+1)=f(1)f(1)=2*2`
`:.f(2)=2^(2) " " "…..(iii)"`
Now, on putting `x=1,y=2` in Eq. (i), we get
`f(1+2)=f(1)f(2)=2*2^(2)" " [" from Eqs. (ii) and (iii) "]`
`:. f(3)=2^(3)`
On, putting `x=1,y=2` in Eq. (i), we get
`f(2+2)=f(2)f(2)=2^(2)*2^(2) " " [" from Eq. (iii) "]`
`:. f(4)=2^(4)`
`vdots" " vdots " " vdots " "`
Similarly, `f(lambda)=2^(lambda),lambda in N`
`:.f(a+k)=2^(a+k),a+k inN`
`:.sum_(k=1)^(n)f(a+k)=16(2^(n)-1) implies sum_(k=1)^(n)2^(a+k)=16(2^(n)-1)`
`implies 2^(n)sum_(k=1)^(n)2^(k)=16(2^(n)-1)`
`implies 2^(n)(2^(1)+2^(2)+2^(3)+"...."2^(n))=16(2^(n)-1)`
`implies 2^(n)*(2(2^(n)-1))/((2-1))=16(2^(n)-1)`
`implies 2^(a+1)=16=2^(4)`
`impliesa+1=4`
`:. a=3`.