A natural functional equation

Algebra Level 4

Find the natural number aa for which k=1nf(a+k)=16(2n1)\displaystyle \sum_{k=1}^n f(a+k)=16(2^n-1) where the function ff satisfies the relation f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) for all natural numbers x,yx,y and furthermore f(1)=2.f(1)=2.

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