In the expansion of `( 1+ x)^(n)` , what is the sum of even binomial coefficient ?

Options

A. `2^(n)`
B. `2^(n-1)`
C. `2^(n+1)`
D. none of these

Answer

To find the sum of the even binomial coefficients in the expansion of \((1 + x)^n\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Binomial Expansion**:
The expansion of \((1 + x)^n\) can be expressed using binomial coefficients:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
where \(\binom{n}{k}\) is the binomial coefficient.

2. **Identify Even and Odd Coefficients**:
In the expansion, the coefficients \(\binom{n}{k}\) for even \(k\) are the even binomial coefficients, and those for odd \(k\) are the odd binomial coefficients.

3. **Evaluate the Expression at Specific Values**:
To find the sum of the even coefficients, we can evaluate the expansion at \(x = 1\) and \(x = -1\):
- When \(x = 1\):
\[ (1 + 1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k} \]
- When \(x = -1\):
\[ (1 - 1)^n = 0 = \sum_{k=0}^{n} \binom{n}{k} (-1)^k \]
This can be rewritten as:
\[ 0 = \sum_{k \text{ even}} \binom{n}{k} - \sum_{k \text{ odd}} \binom{n}{k} \]

4. **Set Up the Equations**:
Let \(E\) be the sum of the even coefficients and \(O\) be the sum of the odd coefficients. From the evaluations, we have:
- \(E + O = 2^n\)
- \(E - O = 0\)

5. **Solve the System of Equations**:
From the second equation, we can express \(E\) in terms of \(O\):
\[ E = O \]
Substituting \(E\) into the first equation gives:
\[ E + E = 2^n \implies 2E = 2^n \implies E = 2^{n-1} \]

6. **Conclusion**:
The sum of the even binomial coefficients in the expansion of \((1 + x)^n\) is:
\[ \boxed{2^{n-1}} \]