Find the sum of the products of first n natural numbers, taken two at a time.

Answer

To find the sum of the products of the first \( n \) natural numbers taken two at a time, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
We need to find the sum of the products of the first \( n \) natural numbers taken two at a time. This means we want to calculate:
\[ S = \sum_{1 \leq i < j \leq n} i \cdot j \]

2. **Using the Formula for Sum of Natural Numbers**:
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^n k = \frac{n(n + 1)}{2} \]

3. **Using the Formula for Sum of Squares**:
The sum of the squares of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

4. **Relating the Sums**:
We can express \( S \) in terms of the square of the sum of the first \( n \) natural numbers:
\[ S = \frac{1}{2} \left( \left( \sum_{k=1}^n k \right)^2 - \sum_{k=1}^n k^2 \right) \]
This is because \( \left( \sum_{k=1}^n k \right)^2 \) includes all products \( i \cdot j \) (for \( i \neq j \)) twice, plus the squares \( k^2 \).

5. **Substituting the Formulas**:
Substitute the formulas for the sums into the equation:
\[ S = \frac{1}{2} \left( \left( \frac{n(n + 1)}{2} \right)^2 - \frac{n(n + 1)(2n + 1)}{6} \right) \]

6. **Simplifying the Expression**:
First, calculate \( \left( \frac{n(n + 1)}{2} \right)^2 \):
\[ \left( \frac{n(n + 1)}{2} \right)^2 = \frac{n^2(n + 1)^2}{4} \]
Now, substitute this back into the equation for \( S \):
\[ S = \frac{1}{2} \left( \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{6} \right) \]

7. **Finding a Common Denominator**:
The common denominator for the two fractions is 12:
\[ S = \frac{1}{2} \left( \frac{3n^2(n + 1)^2}{12} - \frac{2n(n + 1)(2n + 1)}{12} \right) \]
Simplifying gives:
\[ S = \frac{1}{24} \left( 3n^2(n + 1)^2 - 2n(n + 1)(2n + 1) \right) \]

8. **Expanding and Combining Terms**:
Expanding both terms:
- \( 3n^2(n + 1)^2 = 3n^2(n^2 + 2n + 1) = 3n^4 + 6n^3 + 3n^2 \)
- \( 2n(n + 1)(2n + 1) = 2n(2n^2 + 3n + 1) = 4n^3 + 6n^2 + 2n \)

Now, combine:
\[ S = \frac{1}{24} \left( 3n^4 + 6n^3 + 3n^2 - (4n^3 + 6n^2 + 2n) \right) \]
Simplifying further:
\[ S = \frac{1}{24} \left( 3n^4 + 2n^3 - 3n^2 - 2n \right) \]

9. **Final Expression**:
Thus, the sum of the products of the first \( n \) natural numbers taken two at a time is:
\[ S = \frac{n(n + 1)(n - 1)(3n + 2)}{24} \]