Question 76

Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f (a + 2) + ... + f(a + n) = 16 (2$$^n$$ - 1) then a is equal to

Correct Answer: 3

Solution

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

.......=> f(x)=$$2^x$$

Now, f(a + 1) +f (a + 2) + ... + f(a + n) = 16 (2$$^n$$ - 1)

On putting n=1 in the equation we get, f(a+1)=16 => f(a)*f(1)=16 (It is given that f (x + y) = f (x) f (y))

=> $$2^a$$*2=16

=> a=3

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