Find the value of `[( ^10C_1 + ^10C_3 + ^10C_5 + …) xx (20C_1 + ^20C_3 + ^20C_5 + …) xx` `(30C_1 + ^30C_3 + ^30C_5 + …) xx …. xx (100C_1 + ^100C_3 + ^100C_5 + ….)]`

Options

A. A) 540
B. B)`55^2`
C. C) `2^(540)`
D. D) `22^55 - 1`

Answer

To solve the problem, we need to evaluate the expression given, which consists of sums of binomial coefficients. The expression can be broken down into parts, and we will use a known formula for the sum of binomial coefficients at odd indices.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression we need to evaluate is:
\[ ( ^{10}C_1 + ^{10}C_3 + ^{10}C_5 + \ldots) \times (^{20}C_1 + ^{20}C_3 + ^{20}C_5 + \ldots) \times (^{30}C_1 + ^{30}C_3 + ^{30}C_5 + \ldots) \times \ldots \times (^{100}C_1 + ^{100}C_3 + ^{100}C_5 + \ldots) \]

2. **Using the Formula for Odd Indexed Sums**:
The sum of binomial coefficients at odd indices can be calculated using the formula:
\[ \sum_{k \text{ odd}} ^nC_k = 2^{n-1} \]
This means for each \( n \), the sum of binomial coefficients at odd indices equals \( 2^{n-1} \).

3. **Calculating Each Component**:
- For \( n = 10 \):
\[ ^{10}C_1 + ^{10}C_3 + ^{10}C_5 + \ldots = 2^{10-1} = 2^9 \]
- For \( n = 20 \):
\[ ^{20}C_1 + ^{20}C_3 + ^{20}C_5 + \ldots = 2^{20-1} = 2^{19} \]
- For \( n = 30 \):
\[ ^{30}C_1 + ^{30}C_3 + ^{30}C_5 + \ldots = 2^{30-1} = 2^{29} \]
- Continuing this pattern, we find:
- For \( n = 40 \): \( 2^{39} \)
- For \( n = 50 \): \( 2^{49} \)
- For \( n = 60 \): \( 2^{59} \)
- For \( n = 70 \): \( 2^{69} \)
- For \( n = 80 \): \( 2^{79} \)
- For \( n = 90 \): \( 2^{89} \)
- For \( n = 100 \): \( 2^{99} \)

4. **Combining the Results**:
The entire expression can now be rewritten as:
\[ (2^9) \times (2^{19}) \times (2^{29}) \times (2^{39}) \times (2^{49}) \times (2^{59}) \times (2^{69}) \times (2^{79}) \times (2^{89}) \times (2^{99}) \]

5. **Simplifying the Product**:
When multiplying powers of the same base, we add the exponents:
\[ 2^{9 + 19 + 29 + 39 + 49 + 59 + 69 + 79 + 89 + 99} \]

6. **Calculating the Exponent**:
Now, we need to calculate the sum of the exponents:
\[ 9 + 19 + 29 + 39 + 49 + 59 + 69 + 79 + 89 + 99 \]
This is an arithmetic series where:
- First term \( a = 9 \)
- Last term \( l = 99 \)
- Number of terms \( n = 10 \)

The sum of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (a + l) = \frac{10}{2} \times (9 + 99) = 5 \times 108 = 540 \]

7. **Final Result**:
Therefore, the value of the entire expression is:
\[ 2^{540} \]

### Conclusion:
The final answer is:
\[ \boxed{2^{540}} \]