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.`

Explanation

It is given that
`f(x + y) = f (x) f(y) and f(1) = 2`
`f (1 + 1) = f (1) f (1) rArr f(2) = 2^(2), f (1 + 2) = f(1) f(2) rArr f(3) = 2^(3), f (2 + 2) = f (2) f(2) rArr f (4) = 2^(4)`
Similarly `f(k) = 2^(k) and f(a) = 2^(a)`
Hence, `underset(k = 1)overset(n)sum f (a + k) = underset(k =1)overset(n)sum f (a) f(k) = f (a) underset(k =1)overset(n)sumf (k) = 2^(a) underset(k = 1)overset(n)sum 2^(k) = 2^(a) {2^(1) + 2^(2) + ... + 2^(n)}`
`= 2^(a) {(2(2^(n) -1))/(2 -1)} = 2^(a+1) (2^(n) - 1)`
But `underset(k =1)overset(n)sum f (a + k) = 16(2^(n) -1)`
`2^(a + 1) (2^(n) - 1) = 16 (2^(n) -1)`
`:. 2^(a + 1) = 2^(4)`
`:. a + 1 = 4 rArr a = 3`