$\begin{array}{l}Thevalueofnaturalnumber\text{'}a\text{'}forwhich\sum _{k=1}^{n}f(a+k)=16(2^n-1),wherethefunctionsatisfies\\ therelationf(x+y)=f\left(x\right).f\left(y\right)forallnaturalnumberx,yandfurtherf\left(1\right)=2is\end{array}$

$\begin{array}{l}Iff(x+y)=f\left(x\right).f\left(y\right)thenf\left(x\right)=a^x\\ Givenf\left(1\right)=2\\ \Rightarrow a^1=2\Rightarrow a=2\\ \therefore f\left(x\right)=2^x\\ \sum _{k=1}^{n}f(a+k)=16(2^n-1)\\ \Rightarrow 2^{a+1}+2^{a+2}+2^{a+3}+{\mathrm{....2}}^{a+n}=16(2^n-1)\\ \Rightarrow 2^a(2^1+2^2+2^3+\mathrm{...}+2^n)=16(2^n-1)\\ \Rightarrow 2^a\frac{2(2^n-1)}{2-1}=16(2^n-1)\\ \Rightarrow 2^{a+1}=2^4\\ \Rightarrow a+1=4\\ \Rightarrow a=3\end{array}$