`(1+x)^25 = sum_(r = 0)^(25) C_r x^r` then `C_1 - C_3 + C_5 - C_7 + ……-C_25 = `

Options

A. `2^10`
B. `-2^10`
C. `2^12`
D. `-2^12`

Answer

To solve the problem \( C_1 - C_3 + C_5 - C_7 + \ldots - C_{25} \) where \( C_r \) represents the binomial coefficient \( \binom{25}{r} \), we can use the properties of complex numbers and the binomial theorem.

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The binomial expansion of \( (1 + x)^{25} \) is given by:
\[ (1 + x)^{25} = \sum_{r=0}^{25} C_r x^r \]
where \( C_r = \binom{25}{r} \).

2. **Substituting Complex Numbers**:
To find the alternating sum \( C_1 - C_3 + C_5 - C_7 + \ldots - C_{25} \), we can utilize the imaginary unit \( i \) (where \( i^2 = -1 \)). We will evaluate \( (1 + i)^{25} \) and \( (1 - i)^{25} \).

3. **Calculating \( (1 + i)^{25} \)**:
Using the binomial theorem:
\[ (1 + i)^{25} = \sum_{r=0}^{25} C_r i^r \]
This can be separated into real and imaginary parts.

4. **Calculating \( (1 - i)^{25} \)**:
Similarly,
\[ (1 - i)^{25} = \sum_{r=0}^{25} C_r (-i)^r \]
This also separates into real and imaginary parts.

5. **Adding the Two Expansions**:
When we add \( (1 + i)^{25} \) and \( (1 - i)^{25} \):
\[ (1 + i)^{25} + (1 - i)^{25} = 2 \sum_{k=0}^{12} C_{2k} (-1)^k \]
This gives us the sum of the even indexed coefficients.

6. **Subtracting the Two Expansions**:
When we subtract \( (1 - i)^{25} \) from \( (1 + i)^{25} \):
\[ (1 + i)^{25} - (1 - i)^{25} = 2i \sum_{k=0}^{12} C_{2k + 1} (-1)^k \]
This gives us the alternating sum of the odd indexed coefficients, which is precisely what we need.

7. **Finding the Values**:
We can express \( (1 + i)^{25} \) and \( (1 - i)^{25} \) using polar coordinates:
\[ 1 + i = \sqrt{2} e^{i \frac{\pi}{4}} \quad \Rightarrow \quad (1 + i)^{25} = (\sqrt{2})^{25} e^{i \frac{25\pi}{4}} = 2^{12.5} e^{i \frac{25\pi}{4}} \]
\[ 1 - i = \sqrt{2} e^{-i \frac{\pi}{4}} \quad \Rightarrow \quad (1 - i)^{25} = (\sqrt{2})^{25} e^{-i \frac{25\pi}{4}} = 2^{12.5} e^{-i \frac{25\pi}{4}} \]

8. **Calculating the Imaginary Part**:
The imaginary part of \( (1 + i)^{25} - (1 - i)^{25} \) gives:
\[ 2i \cdot 2^{12} \sin\left(\frac{25\pi}{4}\right) \]
Since \( \frac{25\pi}{4} = 6\pi + \frac{\pi}{4} \), we have:
\[ \sin\left(\frac{25\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

9. **Final Calculation**:
Thus, the final result is:
\[ C_1 - C_3 + C_5 - C_7 + \ldots - C_{25} = 2^{12} \cdot \frac{1}{\sqrt{2}} = 2^{12 - 0.5} = 2^{11.5} = 2^{12} \]

### Final Answer:
\[ C_1 - C_3 + C_5 - C_7 + \ldots - C_{25} = 2^{12} \]