Let $f$ be a function such that $f (x + y) = f (x) + f (y)$ for all $x,y \in R$ and $f (1) = 100$. Calculate $\sum_{k = 1}^{10}f (k!)$.
1. Given the functional equation \( f(x + y) = f(x) + f(y) \) for all \( x, y \in \mathbb{R} \), we recognize this as Cauchy's functional equation. One of the well-known solutions to this equation, assuming \( f \) is continuous or measurable, is \( f(x) = kx \) for some constant \( k \).

2. We are also given that \( f(1) = 100 \). Substituting \( x = 1 \) into the general solution \( f(x) = kx \), we get:
   \[
   f(1) = k \cdot 1 = k
   \]
   Therefore, \( k = 100 \).

3. Thus, the function \( f \) can be written as:
   \[
   f(x) = 100x
   \]

4. We need to calculate \( \sum_{k=1}^{10} f(k!) \). Using the function \( f(x) = 100x \), we have:
   \[
   f(k!) = 100k!
   \]

5. Therefore, the sum becomes:
   \[
   \sum_{k=1}^{10} f(k!) = \sum_{k=1}^{10} 100k! = 100 \sum_{k=1}^{10} k!
   \]

6. Now, we calculate the sum of the factorials from 1 to 10:
   \[
   \sum_{k=1}^{10} k! = 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!
   \]
   Calculating each factorial:
   \[
   1! = 1
   \]
   \[
   2! = 2
   \]
   \[
   3! = 6
   \]
   \[
   4! = 24
   \]
   \[
   5! = 120
   \]
   \[
   6! = 720
   \]
   \[
   7! = 5040
   \]
   \[
   8! = 40320
   \]
   \[
   9! = 362880
   \]
   \[
   10! = 3628800
   \]

7. Summing these values:
   \[
   1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 + 3628800 = 4037913
   \]

8. Finally, multiplying by 100:
   \[
   100 \sum_{k=1}^{10} k! = 100 \times 4037913 = 403791300
   \]

The final answer is \(\boxed{403791300}\).