∵f(x+y)=f(x)⋅f(y)+f(x)⋅f(y),∀x,y∈R ....(i) And f(0)=1 ....(ii) Now replace x by zero and y by zero we get $\begin{aligned}
& f(0)=f(0) f(0)+f(0) f(0) \
& 1=f(0)+f(0) \
& \therefore \quad f^{\prime}(0)=\frac{1}{2} ...(iii)
\end{aligned}Nowreplaceybyzeroinequation(i),wegetf(x)=\frac{1}{2} f(x)+f^{\prime}(x)or,\frac{1}{2} f(x)=f^{\prime}(x)then\frac{f^{\prime}(x)}{f(x)}=\frac{1}{2}hence\ln |f(x)|=\frac{x}{2}+cPutx=0,wegetc=0\therefore \quad \ln |f(x)|=\frac{x}{2}Then\sum_{n=1}^{100} \ln (f(\eta))=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots+\frac{100}{2}\right)=\frac{5050}{2}=2525$