If `(1+x)^n=a_0+a_1x+a_2x^2).....+a_nx^n` The sum of the series ` a_(0) +a_(4) +a_(8)+a_(12)+ ……` is

Options

A. `2^(n-1) cos""(npi)/4`
B. `2^(n-2) +2^(n/2-1) cos ""(npi)/4`
C. `2^(n-1)+2^(n/2) sin""(npi)/4`
D. `2^(n-1)sin""(npi)/4`

Answer

To solve the problem, we need to find the sum of the series \( a_0 + a_4 + a_8 + a_{12} + \ldots \) from the expansion of \( (1+x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The coefficients \( a_k \) in the expansion of \( (1+x)^n \) are given by the binomial coefficients:
\[ a_k = \binom{n}{k} \]
Therefore, we can express the sum we want as:
\[ S = a_0 + a_4 + a_8 + a_{12} + \ldots \]

2. **Using Roots of Unity**:
To extract the coefficients \( a_0, a_4, a_8, \ldots \), we can use the roots of unity filter. The sum can be computed using the formula:
\[ S = \frac{1}{4} \left( (1+1)^n + (1+i)^n + (1-i)^n + (1-1)^n \right) \]
Here, \( 1, i, -1, -i \) are the fourth roots of unity.

3. **Calculating Each Term**:
- For \( x = 1 \):
\[ (1+1)^n = 2^n \]
- For \( x = i \):
\[ (1+i)^n = \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^n = 2^{n/2} e^{i \frac{n \pi}{4}} \]
- For \( x = -1 \):
\[ (1-1)^n = 0^n = 0 \]
- For \( x = -i \):
\[ (1-i)^n = \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^n = 2^{n/2} e^{-i \frac{n \pi}{4}} \]

4. **Combining the Results**:
Now, substituting these values into the sum:
\[ S = \frac{1}{4} \left( 2^n + 2^{n/2} e^{i \frac{n \pi}{4}} + 2^{n/2} e^{-i \frac{n \pi}{4}} + 0 \right) \]
The terms \( e^{i \frac{n \pi}{4}} + e^{-i \frac{n \pi}{4}} \) simplify to \( 2 \cos \left( \frac{n \pi}{4} \right) \):
\[ S = \frac{1}{4} \left( 2^n + 2^{n/2} \cdot 2 \cos \left( \frac{n \pi}{4} \right) \right) \]
\[ S = \frac{1}{4} \left( 2^n + 2^{n/2 + 1} \cos \left( \frac{n \pi}{4} \right) \right) \]

5. **Final Result**:
Thus, the sum of the series \( a_0 + a_4 + a_8 + a_{12} + \ldots \) is:
\[ S = \frac{1}{4} \left( 2^n + 2^{(n/2 + 1)} \cos \left( \frac{n \pi}{4} \right) \right) \]