Let f:Y\toY be such that f(1) = 2 and f(x+y) = f(x) f(y) for all natural numbers x and y. If \sum^{n}_{k=1}f(a+k)=16(2^n-1), then a is equal to

KEAM 2018 Math Paper

Question : 89 of 120

Marks: +1, -0

Let f:Y→Y be such that f(1) = 2 and f(x+y) = f(x) f(y) for all natural numbers x and y. If

k=1

f(a+k)=16(2n−1), then a is equal to

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