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

Answer

a=3

Explanation

Let `f(n)=2^(n)` for all positive integers n.
Now, for `n=1,f(1)=2=2!`
`rArr` It is true for `n=1`.
Again, let f(k) is true.
`rArr f(k)=2^(k)," for some "k in N.`
Again, `f(k+1)=f(k)*f(1) " " `[by definition]
`=2^(k)*2 " " `[from induction assumption]
`=2^(k+1)`
Therefore, the result is true for `n=k+1`. Hence, by principle of mathematical induction.
`f(n)=2^(n), AA n in N`
Now, `sum_(k=1)^(n)f(a+k)=sum_(k=1)^(n)f(a)f(k)=f(a) sum_(k=1)^(n)2^(k)`
`=f(a)*(2(2^(n)-1))/(2-1)`
`=2^(a)*2(2^(n)-1)=a^(a+1) (2^(n)-1)`
But ` sum_(k=1)^(n)f(a+k)=16(2^(n)-1)=2^(4)(2^(n)-1)`
Therefore, `a+1=4 rArr a=3`