STATEMENT - 1 : If n is even, `.^(2n)C_(1)+.^(2n)C_(3)+.^(2n)C_(5)+"….."+.^(2n)C_(n-1) = 2^(2n-1)`.
STATEMENT - 2 : `.^(2n)C_(1) + .^(2n)C_(3)+ .^(2n)C_(5) + "……"+ .^(2n)C_(2n-1) = 2^(2n-1)`

Options

A. STATEMENT - 1 is true, STATEMENT - 2 is true and STATEMENT - 2 is correct explanation for STATEMENT - 1.
B. STATEMENT - 1 is true, STATEMENT - 2 is true and STATEMENT - 2 is not correct explanation for STATEMENT - 1.
C. STATEMENT-1 is true, STATEMENT-2 is false
D. STATEMENT-1 is false, STATEMENT-2 is true

Answer

To solve the problem, we need to analyze both statements and verify their correctness.

### Step 1: Understanding Statement 1
Statement 1 claims that if \( n \) is even, then:
\[ \binom{2n}{1} + \binom{2n}{3} + \binom{2n}{5} + \ldots + \binom{2n}{n-1} = 2^{2n-1} \]

### Step 2: Understanding Statement 2
Statement 2 claims:
\[ \binom{2n}{1} + \binom{2n}{3} + \binom{2n}{5} + \ldots + \binom{2n}{2n-1} = 2^{2n-1} \]

### Step 3: Using Binomial Theorem
We will use the binomial theorem to analyze both statements. The binomial theorem states:
\[ (1 + x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k \]

### Step 4: Evaluating for \( x = 1 \)
Substituting \( x = 1 \):
\[ (1 + 1)^{2n} = 2^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} \]

### Step 5: Evaluating for \( x = -1 \)
Now substituting \( x = -1 \):
\[ (1 - 1)^{2n} = 0 = \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \]
This can be separated into even and odd terms:
\[ \sum_{k \text{ even}} \binom{2n}{k} - \sum_{k \text{ odd}} \binom{2n}{k} = 0 \]
This implies:
\[ \sum_{k \text{ even}} \binom{2n}{k} = \sum_{k \text{ odd}} \binom{2n}{k} \]

### Step 6: Relating Even and Odd Sums
Since the sum of all coefficients is \( 2^{2n} \), we can write:
\[ 2 \sum_{k \text{ odd}} \binom{2n}{k} = 2^{2n} \]
Thus,
\[ \sum_{k \text{ odd}} \binom{2n}{k} = 2^{2n-1} \]

### Step 7: Analyzing the Sums
The sum of the odd-indexed binomial coefficients includes all odd indices up to \( 2n-1 \):
\[ \sum_{k=1, k \text{ odd}}^{2n} \binom{2n}{k} = \binom{2n}{1} + \binom{2n}{3} + \ldots + \binom{2n}{2n-1} = 2^{2n-1} \]
This confirms Statement 2.

### Step 8: Verifying Statement 1
For Statement 1, we note:
\[ \binom{2n}{1} + \binom{2n}{3} + \ldots + \binom{2n}{n-1} \]
This sum only includes odd-indexed terms up to \( n-1 \), which is less than \( 2n-1 \). Thus, it does not equal \( 2^{2n-1} \).

### Conclusion
- **Statement 1** is **false** because it does not account for all odd terms up to \( 2n-1 \).
- **Statement 2** is **true** as it correctly sums all odd-indexed terms.

### Final Answer
- Statement 1: False
- Statement 2: True