`(1+x)^(n)=a_(0)+a_(1)x+a_(2)x^(2) +......+a_(n)x^(n)` then
Find the sum of the series ` a_(0) +a_(2)+a_(4) +……`

Options

A. `2^(n)`
B. `2^(n-1)`
C. 2
D. `2^(n -2)`

Answer

To solve the problem, we need to find the sum of the coefficients of the even powers of \(x\) in the expansion of \((1+x)^n\). The coefficients are represented as \(a_0, a_2, a_4, \ldots\).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The expansion of \((1+x)^n\) is given by:
\[ (1+x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \]
where \(a_k = \binom{n}{k}\) for \(k = 0, 1, 2, \ldots, n\).

2. **Finding the Sum of Even Coefficients**:
We want to find \(S = a_0 + a_2 + a_4 + \ldots\).

3. **Substituting \(x = 1\)**:
First, substitute \(x = 1\) into the expansion:
\[ (1+1)^n = 2^n \]
This gives us the sum of all coefficients:
\[ S_{\text{total}} = a_0 + a_1 + a_2 + \ldots + a_n = 2^n \]

4. **Substituting \(x = -1\)**:
Next, substitute \(x = -1\):
\[ (1-1)^n = 0 \]
This gives us:
\[ S_{\text{total}} - (a_1 + a_3 + a_5 + \ldots) = 0 \]
or,
\[ S_{\text{total}} = a_0 + a_2 + a_4 + \ldots - (a_1 + a_3 + a_5 + \ldots) \]
Therefore, we can express this as:
\[ 2^n = S + (a_1 + a_3 + a_5 + \ldots) \]

5. **Combining the Two Equations**:
Let \(E\) be the sum of even coefficients \(S\) and \(O\) be the sum of odd coefficients:
\[ E + O = 2^n \quad \text{(1)} \]
\[ E - O = 0 \quad \text{(2)} \]

6. **Solving the System of Equations**:
From equation (2), we have:
\[ E = O \]
Substitute \(O\) in equation (1):
\[ E + E = 2^n \implies 2E = 2^n \implies E = \frac{2^n}{2} = 2^{n-1} \]

7. **Conclusion**:
Thus, the sum of the series \(a_0 + a_2 + a_4 + \ldots\) is:
\[ \boxed{2^{n-1}} \]