Find the sum of the product of the integers `1,2,3…..n` taken two at a time and over the sum of their squares.

Answer

To solve the given problem of finding the sum of the product of the integers \(1, 2, 3, \ldots, n\) taken two at a time and over the sum of their squares, we will follow these steps:

### Step 1: Understand the Problem
We need to find the sum of the products of integers taken two at a time, which can be represented mathematically as:
\[ S = \sum_{1 \leq i < j \leq n} ij \]
We also need to consider the sum of their squares:
\[ Q = \sum_{k=1}^{n} k^2 \]

### Step 2: Calculate the Sum of Products
The sum \(S\) can be rewritten using the formula for combinations:
\[ S = \frac{1}{2} \left( \sum_{i=1}^{n} i \right)^2 - \frac{1}{2} \sum_{i=1}^{n} i^2 \]
Here, \(\sum_{i=1}^{n} i\) is the sum of the first \(n\) integers, which is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]
Thus,
\[ S = \frac{1}{2} \left( \frac{n(n+1)}{2} \right)^2 - \frac{1}{2} \sum_{i=1}^{n} i^2 \]

### Step 3: Calculate the Sum of Squares
The sum of squares \(\sum_{i=1}^{n} i^2\) is given by:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Substitute Values into the Expression for S
Now substituting the values into the expression for \(S\):
\[ S = \frac{1}{2} \left( \frac{n(n+1)}{2} \right)^2 - \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 5: Simplify the Expression
Calculating the first term:
\[ \frac{1}{2} \left( \frac{n(n+1)}{2} \right)^2 = \frac{1}{2} \cdot \frac{n^2(n+1)^2}{4} = \frac{n^2(n+1)^2}{8} \]
Calculating the second term:
\[ \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{12} \]
Thus, we have:
\[ S = \frac{n^2(n+1)^2}{8} - \frac{n(n+1)(2n+1)}{12} \]

### Step 6: Find a Common Denominator and Combine
The common denominator of 8 and 12 is 24. Thus, we rewrite:
\[ S = \frac{3n^2(n+1)^2}{24} - \frac{2n(n+1)(2n+1)}{24} \]
Combining these fractions gives:
\[ S = \frac{3n^2(n+1)^2 - 2n(n+1)(2n+1)}{24} \]

### Step 7: Final Simplification
Now we can simplify the numerator:
\[ 3n^2(n+1)^2 - 2n(n+1)(2n+1) = n(n+1) \left( 3n(n+1) - 2(2n+1) \right) \]
Expanding this:
\[ = n(n+1)(3n^2 + 3n - 4n - 2) = n(n+1)(3n^2 - n - 2) \]
Thus,
\[ S = \frac{n(n+1)(3n^2 - n - 2)}{24} \]

### Step 8: Conclusion
The sum of the product of the integers \(1, 2, 3, \ldots, n\) taken two at a time is:
\[ S = \frac{n(n+1)(3n^2 - n - 2)}{24} \]