The continued product `2.6.10.14…..`(`n` times) in equal to

Options

A. `.^(2n)C_(n)`
B. `.^(2n)P_(n)`
C. `(n + 1)(n + 2)(n + 3) …….(n + n)`
D. `4^(n)`

Explanation

To solve the problem of finding the continued product \(2 \times 6 \times 10 \times 14 \ldots\) (up to \(n\) times), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Pattern**:
The terms in the product are \(2, 6, 10, 14, \ldots\). We can observe that these terms can be expressed in the form:
\[ a_k = 2 + 4(k - 1) = 4k - 2 \]
where \(k\) ranges from \(1\) to \(n\). Thus, the \(k\)-th term is \(4k - 2\).

2. **Write the Product**:
The continued product can be written as:
\[ P_n = \prod_{k=1}^{n} (4k - 2) \]
This expands to:
\[ P_n = (4 \cdot 1 - 2)(4 \cdot 2 - 2)(4 \cdot 3 - 2) \ldots (4n - 2) \]

3. **Factor Out Common Terms**:
We can factor out \(2\) from each term:
\[ P_n = 2^n \prod_{k=1}^{n} (2k - 1) \]
Here, \(2k - 1\) represents the sequence of odd numbers.

4. **Express the Product of Odd Numbers**:
The product of the first \(n\) odd numbers can be expressed as:
\[ \prod_{k=1}^{n} (2k - 1) = \frac{(2n)!}{2^n n!} \]
This is a known result.

5. **Combine the Results**:
Now substituting this back into our expression for \(P_n\):
\[ P_n = 2^n \cdot \frac{(2n)!}{2^n n!} = \frac{(2n)!}{n!} \]

6. **Final Result**:
Therefore, the continued product \(2 \times 6 \times 10 \times 14 \ldots\) (up to \(n\) times) is:
\[ P_n = \frac{(2n)!}{n!} \]