Evaluate the `""^16C_1 + ""^16C_3 +""^16C_5+…………..+""^16C_15`

Answer

To evaluate the expression \( \binom{16}{1} + \binom{16}{3} + \binom{16}{5} + \ldots + \binom{16}{15} \), we can use properties of binomial coefficients and the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Sum of Odd Indexed Binomial Coefficients**:
The sum \( \binom{16}{1} + \binom{16}{3} + \binom{16}{5} + \ldots + \binom{16}{15} \) includes all the odd indexed binomial coefficients from \( n = 16 \).

2. **Using the Binomial Theorem**:
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
For \( n = 16 \), we can set \( x = 1 \) and \( y = 1 \):
\[ (1 + 1)^{16} = \sum_{k=0}^{16} \binom{16}{k} = 2^{16} \]

3. **Separating Even and Odd Indexed Terms**:
We can also evaluate \( (1 - 1)^{16} \):
\[ (1 - 1)^{16} = \sum_{k=0}^{16} \binom{16}{k} (-1)^k = 0 \]
This implies that:
\[ \sum_{k \text{ even}} \binom{16}{k} - \sum_{k \text{ odd}} \binom{16}{k} = 0 \]
Let \( A = \sum_{k \text{ even}} \binom{16}{k} \) and \( B = \sum_{k \text{ odd}} \binom{16}{k} \). Therefore, we have:
\[ A - B = 0 \implies A = B \]

4. **Finding the Value of A and B**:
From the first equation, we know:
\[ A + B = 2^{16} \]
Since \( A = B \), we can substitute:
\[ 2A = 2^{16} \implies A = 2^{15} \]
Thus, \( B = 2^{15} \).

5. **Conclusion**:
Therefore, the sum of the odd indexed binomial coefficients is:
\[ \binom{16}{1} + \binom{16}{3} + \binom{16}{5} + \ldots + \binom{16}{15} = 2^{15} \]

### Final Answer:
\[ \binom{16}{1} + \binom{16}{3} + \binom{16}{5} + \ldots + \binom{16}{15} = 32768 \]