If `S_(n) = (.^(n)C_(0))^(2) + (.^(n)C_(1))^(2) + (.^(n)C_(n))^(n)`, then maximum value of `[(S_(n+1))/(S_(n))]` is `"_____"`.
(where `[*]` denotes the greatest integer function)

Explanation

To solve the problem, we need to find the maximum value of \(\left\lfloor \frac{S_{n+1}}{S_n} \right\rfloor\), where \(S_n = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \ldots + \binom{n}{n}^2\).

### Step 1: Express \(S_n\) in a simplified form
Using the identity for the sum of squares of binomial coefficients, we know that:
\[ S_n = \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]
Thus, we can write:
\[ S_n = \binom{2n}{n} \]

### Step 2: Write \(S_{n+1}\)
Now, we can express \(S_{n+1}\) similarly:
\[ S_{n+1} = \binom{2(n+1)}{n+1} = \binom{2n + 2}{n + 1} \]

### Step 3: Set up the ratio \(\frac{S_{n+1}}{S_n}\)
Now we need to find:
\[ \frac{S_{n+1}}{S_n} = \frac{\binom{2n + 2}{n + 1}}{\binom{2n}{n}} \]

### Step 4: Use the formula for binomial coefficients
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
we can write:
\[ \frac{S_{n+1}}{S_n} = \frac{\frac{(2n + 2)!}{(n + 1)!(n + 1)!}}{\frac{(2n)!}{n!n!}} = \frac{(2n + 2)! \cdot n! \cdot n!}{(n + 1)! \cdot (n + 1)! \cdot (2n)!} \]

### Step 5: Simplify the ratio
This simplifies to:
\[ \frac{S_{n+1}}{S_n} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{2(2n + 1)}{n + 1} \]

### Step 6: Further simplify
This can be rewritten as:
\[ \frac{S_{n+1}}{S_n} = 4 - \frac{2}{n + 1} \]

### Step 7: Find the maximum value of \(\left\lfloor \frac{S_{n+1}}{S_n} \right\rfloor\)
As \(n\) approaches infinity, \(\frac{2}{n + 1}\) approaches 0, thus:
\[ \frac{S_{n+1}}{S_n} \to 4 \]
The greatest integer function gives us:
\[ \left\lfloor \frac{S_{n+1}}{S_n} \right\rfloor \to \left\lfloor 4 - \frac{2}{n + 1} \right\rfloor \]
For large \(n\), this approaches 3.

### Conclusion
Thus, the maximum value of \(\left\lfloor \frac{S_{n+1}}{S_n} \right\rfloor\) is:
\[ \boxed{3} \]