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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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