Evaluate :
`1+^(15)C_(1)+^(15)C_(2)+^(15)C_(3)+......+^(15)C_(15)`

Explanation

To evaluate the expression \(1 + \binom{15}{1} + \binom{15}{2} + \binom{15}{3} + \ldots + \binom{15}{15}\), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**: The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, we can set \(a = 1\) and \(b = 1\), and \(n = 15\).

2. **Applying the Binomial Theorem**: By substituting \(a\) and \(b\) into the theorem, we have:
\[ (1 + 1)^{15} = \sum_{k=0}^{15} \binom{15}{k} 1^{15-k} 1^k \]
This simplifies to:
\[ 2^{15} = \sum_{k=0}^{15} \binom{15}{k} \]

3. **Identifying the Terms**: The sum on the right-hand side includes all the binomial coefficients from \(k = 0\) to \(k = 15\):
\[ \sum_{k=0}^{15} \binom{15}{k} = \binom{15}{0} + \binom{15}{1} + \binom{15}{2} + \ldots + \binom{15}{15} \]

4. **Separating the First Term**: We can separate the first term from the sum:
\[ 1 + \binom{15}{1} + \binom{15}{2} + \ldots + \binom{15}{15} = \sum_{k=0}^{15} \binom{15}{k} \]

5. **Final Calculation**: Thus, we have:
\[ 1 + \binom{15}{1} + \binom{15}{2} + \ldots + \binom{15}{15} = 2^{15} \]

6. **Evaluating \(2^{15}\)**: Now, we calculate \(2^{15}\):
\[ 2^{15} = 32768 \]

### Conclusion:
The value of the expression \(1 + \binom{15}{1} + \binom{15}{2} + \ldots + \binom{15}{15}\) is \(32768\).