For integers n and r, let `([n], [r])={(""^nC_r," ""if " n ge r ge 0), (0," ""otherwise"):}`. The maximum value of k for which the sum `sum_(i=0)^(k)([10], [i])([15], [k-i])+sum_(i=0)^(k+1)([12], [i])([13], [k+1-i])`exists, is equal to _________.

Answer

To solve the problem, we need to analyze the given sums and determine the maximum value of \( k \) for which the sums exist.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
We have a function defined as:
\[ [n, r] = \begin{cases} \binom{n}{r} & \text{if } n \geq r \geq 0 \\ 0 & \text{otherwise} \end{cases} \]
This means that \( [n, r] \) is the binomial coefficient \( \binom{n}{r} \) when \( n \) is greater than or equal to \( r \) and both are non-negative integers. Otherwise, it is zero.

2. **Setting Up the Sums**:
We need to evaluate the following sums:
\[ S(k) = \sum_{i=0}^{k} [10, i][15, k-i] + \sum_{i=0}^{k+1} [12, i][13, k+1-i] \]

3. **Evaluating the First Sum**:
The first sum can be interpreted using the binomial theorem:
\[ \sum_{i=0}^{k} [10, i][15, k-i] = \binom{10 + 15}{k} = \binom{25}{k} \]
This holds true as long as \( k \leq 10 \) and \( k \leq 15 \).

4. **Evaluating the Second Sum**:
Similarly, for the second sum:
\[ \sum_{i=0}^{k+1} [12, i][13, k+1-i] = \binom{12 + 13}{k+1} = \binom{25}{k+1} \]
This holds true as long as \( k+1 \leq 12 \) and \( k+1 \leq 13 \).

5. **Finding the Maximum \( k \)**:
For the first sum, the maximum \( k \) is \( 10 \) (since \( k \) must be less than or equal to \( 10 \)).
For the second sum, the maximum \( k+1 \) is \( 12 \) (which means \( k \) can be at most \( 11 \)).

6. **Determining the Overall Maximum**:
To satisfy both conditions:
- From the first sum, \( k \leq 10 \)
- From the second sum, \( k \leq 11 \)

The more restrictive condition is \( k \leq 10 \).

Thus, the maximum value of \( k \) for which the sum exists is:

\[ \boxed{10} \]