The sum of all the products of the first n(+) ive integers taken two at a time is

Options

A. `(1)/(48) (n - 2) (n - 1) n^(2)`
B. `(1)/(6) (n - 1) n (n + 1) (3n + 2)`
C. `(1)/(6) n(n + 1) (n + 2) (n + 5)`
D. none

Answer

To find the sum of all the products of the first n positive integers taken two at a time, we can use the following steps:

### Step 1: Understand the Problem
We need to find the sum of the products of the first n positive integers taken two at a time. This means we want to calculate the sum of all possible products of pairs of integers from the set {1, 2, ..., n}.

### Step 2: Use the Formula for the Sum of Natural Numbers
The sum of the first n natural numbers is given by the formula:
\[ S = \frac{n(n + 1)}{2} \]

### Step 3: Use the Formula for the Sum of Squares
The sum of the squares of the first n natural numbers is given by:
\[ S_{squares} = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Apply the Binomial Theorem
According to the binomial theorem, we have:
\[ \left( \sum_{i=1}^{n} x_i \right)^2 = \sum_{i=1}^{n} x_i^2 + 2 \sum_{1 \leq i < j \leq n} x_i x_j \]
where \( \sum_{1 \leq i < j \leq n} x_i x_j \) represents the sum of the products taken two at a time.

### Step 5: Set Up the Equation
Let \( S = \sum_{i=1}^{n} i \). Then we can write:
\[ S^2 = \sum_{i=1}^{n} i^2 + 2 \sum_{1 \leq i < j \leq n} ij \]
From this, we can isolate the term we want:
\[ 2 \sum_{1 \leq i < j \leq n} ij = S^2 - \sum_{i=1}^{n} i^2 \]

### Step 6: Substitute the Values
Substituting the values of \( S \) and \( S_{squares} \):
\[ 2 \sum_{1 \leq i < j \leq n} ij = \left( \frac{n(n + 1)}{2} \right)^2 - \frac{n(n + 1)(2n + 1)}{6} \]

### Step 7: Simplify the Expression
Now we simplify the right-hand side:
1. Calculate \( S^2 \):
\[ S^2 = \left( \frac{n(n + 1)}{2} \right)^2 = \frac{n^2(n + 1)^2}{4} \]
2. Substitute this into the equation:
\[ 2 \sum_{1 \leq i < j \leq n} ij = \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{6} \]

### Step 8: Find a Common Denominator
To combine the fractions, find a common denominator (which is 12):
\[ 2 \sum_{1 \leq i < j \leq n} ij = \frac{3n^2(n + 1)^2 - 2n(n + 1)(2n + 1)}{12} \]

### Step 9: Factor and Simplify
Now we need to simplify the numerator:
\[ 3n^2(n + 1)^2 - 2n(n + 1)(2n + 1) \]
This will yield the final expression for the sum of products taken two at a time.

### Step 10: Divide by 2
Finally, to find the sum of products taken two at a time, divide the result by 2:
\[ \sum_{1 \leq i < j \leq n} ij = \frac{1}{2} \left( \frac{3n^2(n + 1)^2 - 2n(n + 1)(2n + 1)}{12} \right) \]

### Final Result
After simplifying, we will arrive at the final answer.

---