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:

Step 1: Analyze the given functional equation

We are given a functional equation $f(x + y) = f(x) f'(y) + f'(x) f(y)$ and the initial condition $f(0) = 1$. To simplify the functional equation, we substitute $x=0$ into the equation. This allows us to relate $f(y)$ and $f'(y)$ using the given initial condition.

Step 2: Solve the differential equation

Rearranging the equation from the previous step, we get a first-order linear differential equation. Let $k = 1 - f'(0)$. Then the equation becomes $f'(y) = k f(y)$. This is a standard separable differential equation.

Step 3: Integrate to find f(y)

Integrating both sides of the differential equation $\frac{f'(y)}{f(y)} = k$ with respect to $y$, we get $\log_e |f(y)| = ky + C$. Exponentiating both sides gives $f(y) = e^{ky+C} = e^C e^{ky}$. Let $A = e^C$. So, $f(y) = A e^{ky}$.

Step 4: Apply initial conditions to find constants

Using the initial condition $f(0) = 1$, we find that $A=1$. So, $f(y) = e^{ky}$. Now we need to find the value of $k$. We know $k = 1 - f'(0)$. Differentiating $f(y) = e^{ky}$ gives $f'(y) = k e^{ky}$. Substituting these back into the simplified functional equation $f(y) = f'(y) + f'(0)f(y)$ leads to $1 = k + f'(0)$, which is consistent with our definition of $k$. This means $k$ can be any real constant, and $f'(0)$ is determined by $k$. The problem statement implies a unique function, so we must re-examine the original functional equation.

Step 5: Re-evaluate using the original functional equation

Let's substitute $f(x) = e^{kx}$ and $f'(x) = k e^{kx}$ into the original functional equation $f(x + y) = f(x) f'(y) + f'(x) f(y)$. We find that $e^{kx}e^{ky} = 2k e^{kx}e^{ky}$. This implies $1 = 2k$. Therefore, $k = \frac{1}{2}$.

Step 6: Determine f(n) and calculate the sum

With $k = \frac{1}{2}$, the function is $f(x) = e^{x/2}$. Now we need to calculate $\log_e f(n)$, which simplifies to $\frac{n}{2}$. The sum $\sum_{n=1}^{100} \log_e f(n)$ becomes $\sum_{n=1}^{100} \frac{n}{2}$. This is an arithmetic series, and we can use the formula for the sum of the first $N$ natural numbers, $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$.

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