If f is a function such that $\sum _{k=1}^{n}f(a+k)=16(2^n-1),$$f(x+y)=f\left(x\right)f\left(y\right)\forall x,y\in N$  and f (1) = 2 then a =

$$\begin{gathered} f(x+y)=f\left(x\right)f\left(y\right) \\ f\left(x\right)=a^x\Rightarrow f\left(1\right)=a=2 \\ f\left(2\right)=f(1+1)=f\left(1\right)f\left(1\right)=2^2 \\ f\left(3\right)=f(2+1)=f\left(2\right)f\left(1\right)=2^3 \\ \sum _{k=1}^{n}f(a+k)=16(2^n-1) \\ f(a+1)+f(a+2)+f(a+3)+\mathrm{....}+f(a+n)=16(2^n-1) \\ f\left(a\right)(2+2^2+2^3+\mathrm{...}+2^n)=16(2^n-1) \\ 2f\left(a\right)(1+2+2^2+2^3\mathrm{....})=16(2^n-1)=8(2^n-1) \\ {2}^a=2^3\Rightarrow a=3 \end{gathered}$$