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

Given the functional equation $$ f(x+y) = f(x) f'(y) + f'(x) f(y) $$ for all real $$ x $$ and $$ y $$, and the initial condition $$ f(0) = 1 $$, we need to find $$ \sum_{n=1}^{100} \log_e f(n) $$.

To solve, substitute $$ y = 0 $$ into the functional equation:

$$ f(x+0) = f(x) f'(0) + f'(x) f(0) $$

Given $$ f(0) = 1 $$, this simplifies to:

$$ f(x) = f(x) f'(0) + f'(x) $$

Rearranging terms:

$$ f'(x) = f(x) - f(x) f'(0) $$

$$ f'(x) = f(x) (1 - f'(0)) $$

Let $$ k = 1 - f'(0) $$, a constant. Then:

$$ f'(x) = k f(x) $$

This is a first-order linear differential equation. The general solution is $$ f(x) = A e^{k x} $$, where $$ A $$ is a constant.

Using the initial condition $$ f(0) = 1 $$:

$$ f(0) = A e^{k \cdot 0} = A = 1 $$

So, $$ f(x) = e^{k x} $$.

Now, find $$ f'(x) $$:

$$ f'(x) = k e^{k x} $$

Evaluate at $$ x = 0 $$:

$$ f'(0) = k e^{0} = k $$

But $$ k = 1 - f'(0) $$, so substitute:

$$ k = 1 - k $$

$$ 2k = 1 $$

$$ k = \frac{1}{2} $$

Thus, $$ f(x) = e^{\frac{x}{2}} $$.

Verify this satisfies the original functional equation:

Left-hand side: $$ f(x+y) = e^{\frac{x+y}{2}} $$.

Right-hand side: $$ f(x) f'(y) + f'(x) f(y) = e^{\frac{x}{2}} \cdot \frac{1}{2} e^{\frac{y}{2}} + \frac{1}{2} e^{\frac{x}{2}} \cdot e^{\frac{y}{2}} = \frac{1}{2} e^{\frac{x+y}{2}} + \frac{1}{2} e^{\frac{x+y}{2}} = e^{\frac{x+y}{2}} $$.

Both sides are equal, so the solution is valid.

Now compute $$ \log_e f(n) $$:

$$ \log_e f(n) = \log_e \left( e^{\frac{n}{2}} \right) = \frac{n}{2} $$

The sum is:

$$ \sum_{n=1}^{100} \log_e f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \sum_{n=1}^{100} n $$

The sum of the first $$ n $$ natural numbers is given by the formula $$ \sum_{n=1}^{m} n = \frac{m(m+1)}{2} $$. For $$ m = 100 $$:

$$ \sum_{n=1}^{100} n = \frac{100 \times 101}{2} = 5050 $$

Therefore,

$$ \frac{1}{2} \times 5050 = 2525 $$

The sum $$ \sum_{n=1}^{100} \log_e f(n) = 2525 $$.

Comparing with the options, A. 2525 is correct.