The value of `sum_(r=0)^nsum_(s=0)^nsum_(t=0)^nsum_(u=0)^n(1)` is

Options

A. `""^(n)C_(4)`
B. `""^(n+1)C_(4)`
C. `""^(4n+1)C_(4)`
D. `(n+1)^(4)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{r=0}^{n} \sum_{s=0}^{n} \sum_{t=0}^{n} \sum_{u=0}^{n} 1 \]

### Step-by-Step Solution:

1. **Understanding the Summation**:
The expression \(\sum_{r=0}^{n} \sum_{s=0}^{n} \sum_{t=0}^{n} \sum_{u=0}^{n} 1\) represents the total number of combinations of \(r\), \(s\), \(t\), and \(u\) where each variable ranges from 0 to \(n\).

2. **Counting the Terms**:
Each summation \(\sum_{x=0}^{n} 1\) counts the number of terms from \(x=0\) to \(x=n\). There are \(n + 1\) terms in each summation (including 0).

3. **Calculating the Total**:
Since there are four independent summations (for \(r\), \(s\), \(t\), and \(u\)), the total number of combinations can be calculated as:
\[ (n + 1) \times (n + 1) \times (n + 1) \times (n + 1) = (n + 1)^4 \]

4. **Final Result**:
Therefore, the value of the original expression is:
\[ (n + 1)^4 \]

### Conclusion:
The value of \(\sum_{r=0}^{n} \sum_{s=0}^{n} \sum_{t=0}^{n} \sum_{u=0}^{n} 1\) is \((n + 1)^4\).