The sum of the products of 2n numbers `pm1,pm2,pm3, . . . . ,n` taking two at time is

Options

A. `-underset(r=1)overset(n)sumr`
B. `underset(r=1)overset(n)sumr^(2)`
C. `-underset(r=1)overset(n)sumr^(2)`
D. none of these

Explanation

To solve the problem of finding the sum of the products of \(2n\) numbers \(pm_1, pm_2, pm_3, \ldots, pm_n\) taking two at a time, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Numbers**: The numbers we are dealing with are \(pm_1, pm_2, pm_3, \ldots, pm_n\) where \(pm_i\) can take values of \(-i\) and \(+i\) for \(i = 1, 2, \ldots, n\). This means we have the numbers:
\[ -1, 1, -2, 2, -3, 3, \ldots, -n, n \]

2. **Identifying the Products**: We need to calculate the sum of the products of these numbers taken two at a time. The products can be represented as:
\[ (-i)(-j) + (-i)(j) + (i)(-j) + (i)(j) \]
for \(i \neq j\).

3. **Calculating the Products**: The products can be simplified as follows:
- The product of two negative numbers gives a positive product.
- The product of one negative and one positive number gives a negative product.
- The product of two positive numbers gives a positive product.

4. **Summing the Products**: The sum of all products taken two at a time can be expressed as:
\[ S = \sum_{1 \leq i < j \leq n} (pm_i \cdot pm_j) \]

5. **Using the Formula for Sums**: The sum of the products can be calculated using the formula for the sum of squares. The sum of squares of the first \(n\) natural numbers is given by:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

6. **Final Calculation**: Since we are interested in the sum of products of \(2n\) numbers, we can express this as:
\[ S = -\left(\sum_{r=1}^{n} r^2\right) = -\frac{n(n + 1)(2n + 1)}{6} \]

### Final Answer:
Thus, the sum of the products of the \(2n\) numbers taken two at a time is:
\[ -\frac{n(n + 1)(2n + 1)}{6} \]