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

`sum_(k=1)^n​f(a+k)=16(2^n−1) `


`f(a).f(1)+f(a).f(2)+...+f(a).f(n)
=16(2^n−1) `

`f(a).[f(1)+f(2)+...+f(n)]=16(2^n−1) `

Also, `f(2)=f(1+1)=f(1).f(1)=[f(1)]^2=4`

Similarly, `f(3)=f(2+1)=f(2).f(1)=4×2=8`

`f(a).[f(1)+f(2)+...+f(n)]=16(2^n−1)`
becomes

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