$\text{ Let }\sum _{k=1}^{10}f(a+K)=16\left(2^{10}-1\right),\text{ where the function }f\left(x\right)\text{ satisfies }f(x+y)=f\left(x\right)f\left(y\right)\text{ for all }$$\text{ natural numbers }x,y\text{ and }f\left(1\right)=2,\text{ then the natural number ' }\mathrm{a}\text{ ' is }$

$f\left(x+y\right)=f\left(x\right)f\left(y\right)\Rightarrow f\left(x\right)=2^x$   $\left(\because f\left(1\right)=2\right)$

$\sum _{k=1}^{10}f\left(a+k\right)=f\left(a+1\right)+f\left(a+2\right)+\mathrm{.....}+f\left(a+10\right)$

                    =$2^a\left(2+2^2+\mathrm{.......}+2^{10}\right)$

                   =$2^a\left(2\left(2^{10}-1\right)\right)$

                   =$2^{a+1}\left(2^{10}-1\right)$