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(B) Given $f(x+y)=f(x)f'(y)+f'(x)f(y)$.Setting $x=0$ and $y=0$,we get $f(0)=f(0)f'(0)+f'(0)f(0)$,which implies $1=2f'(0)$,so $f'(0)=\frac{1}{2}$.Setti

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$2525$

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(B) Given $f(x+y)=f(x)f'(y)+f'(x)f(y)$.Setting $x=0$ and $y=0$,we get $f(0)=f(0)f'(0)+f'(0)f(0)$,which implies $1=2f'(0)$,so $f'(0)=\frac{1}{2}$.Setting $y=0$ in the original equation,we get $f(x)=f(x)f'(0)+f'(x)f(0)$.Substituting $f(0)=1$ and $f'(0)=\frac{1}{2}$,we have $f(x)=\frac{1}{2}f(x)+f'(x)$,which simplifies to $f'(x)=\frac{f(x)}{2}$.This is a separable differential equation: $\frac{f'(x)}{f(x)}=\frac{1}{2}$.Integrating both sides with respect to $x$,we get $\ln|f(x)|=\frac{x}{2}+C$.Since $f(0)=1$,we have $\ln(1)=0+C$,so $C=0$.Thus,$\ln f(x)=\frac{x}{2}$,which means $f(x)=e^{x/2}$.Now,$\sum_{n=1}^{100} \log_{e} f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} 	imes \frac{100(101)}{2} = \frac{5050}{2} = 2525$.

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x)f'(y)+f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\sum_{n=1}^{100} \log_{e} f(n)$ is equal to:

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