Prove that `sum_(r=0)^(5) ""^(5)C_(r )=32`.

Answer

To prove that \(\sum_{r=0}^{5} \binom{5}{r} = 32\), we can use the binomial theorem.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The binomial theorem states that:
\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r \]
where \(\binom{n}{r}\) is the binomial coefficient.

2. **Setting Up the Equation**:
For our case, we will set \(n = 5\), \(x = 1\), and \(y = 1\). Thus, we have:
\[ (1 + 1)^5 = \sum_{r=0}^{5} \binom{5}{r} 1^{5-r} 1^r \]

3. **Simplifying the Left Side**:
The left side simplifies to:
\[ 2^5 \]

4. **Equating Both Sides**:
Therefore, we can write:
\[ 2^5 = \sum_{r=0}^{5} \binom{5}{r} \]

5. **Calculating \(2^5\)**:
Now, calculating \(2^5\):
\[ 2^5 = 32 \]

6. **Conclusion**:
Thus, we have:
\[ \sum_{r=0}^{5} \binom{5}{r} = 32 \]
which proves the statement.