Solution:

To find the value of the expression 2nC2, 2nC4, ..., 2nC2n, we need to understand the concept of combinations.

A combination is a selection of items from a larger set without regard to the order of the items. In other words, it is the number of ways to choose a certain number of items from a set.

The general formula for finding the number of combinations, denoted as nCr, is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of items in the set and r is the number of items to be selected.

In the given expression, we have 2nC2, 2nC4, ..., 2nC2n. Let's break down each term:

1. 2nC2:
- The number of items in the set is 2n.
- We want to select 2 items from the set.
- Using the combination formula, we have 2n! / (2!(2n-2)!) = (2n * (2n-1)) / 2 = n * (2n-1).

2. 2nC4:
- The number of items in the set is 2n.
- We want to select 4 items from the set.
- Using the combination formula, we have 2n! / (4!(2n-4)!) = (2n * (2n-1) * (2n-2) * (2n-3)) / (4 * 3 * 2 * 1) = (n * (2n-1) * (2n-2) * (2n-3)) / 6.

3. 2nC6:
- The number of items in the set is 2n.
- We want to select 6 items from the set.
- Using the combination formula, we have 2n! / (6!(2n-6)!) = (2n * (2n-1) * (2n-2) * (2n-3) * (2n-4) * (2n-5)) / (6 * 5 * 4 * 3 * 2 * 1) = (n * (2n-1) * (2n-2) * (2n-3) * (2n-4) * (2n-5)) / 720.

By observing the pattern, we can see that the denominator of each term is always a factorial of the number of items to be selected (r!). Therefore, the general form of the expression can be written as:

2nCr = (n * (2n-1) * (2n-2) * ... * (2n-r+1)) / r!

In the given expression, we have 2nC2, 2nC4, ..., 2nC2n. The value of the expression can be written as:

2nC2 * 2nC4 * ... * 2nC2n = (n * (2n-1)) * (n * (2n-1) * (2n-2) * (2n-3)) * ... * (n * (2n-1) *