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

### Step 1: Define the sum of the products of integers taken two at a time
The sum of the product of integers taken two at a time can be expressed as:
\[ S_1 = \sum_{k=1}^{n} k(k+1) \]
This represents the sum of products \(1 \times 2, 2 \times 3, \ldots, (n-1) \times n\).

### Step 2: Expand the summation
We can expand \(S_1\) as follows:
\[ S_1 = \sum_{k=1}^{n} (k^2 + k) \]
This can be separated into two summations:
\[ S_1 = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 3: Use known formulas for summations
We can use the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
- The sum of the first \(n\) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
- The sum of the squares of the first \(n\) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Substitute the formulas into \(S_1\)
Substituting these formulas into our expression for \(S_1\):
\[ S_1 = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]

### Step 5: Simplify \(S_1\)
To combine the two terms, we need a common denominator:
\[ S_1 = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} \]
This gives:
\[ S_1 = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n+4)}{6} \]
Factoring out the common terms:
\[ S_1 = \frac{n(n+1)(n+2)}{3} \]

### Step 6: Define the sum of squares
Next, we define the sum of squares \(S_2\):
\[ S_2 = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 7: Find the ratio \( \frac{S_1}{S_2} \)
Now we need to find the ratio \( \frac{S_1}{S_2} \):
\[ \frac{S_1}{S_2} = \frac{\frac{n(n+1)(n+2)}{3}}{\frac{n(n+1)(2n+1)}{6}} \]
This simplifies to:
\[ \frac{S_1}{S_2} = \frac{n(n+1)(n+2) \cdot 6}{3 \cdot n(n+1)(2n+1)} = \frac{2(n+2)}{2n+1} \]

### Final Answer
Thus, the required sum of the product of the integers taken two at a time over the sum of their squares is:
\[ \frac{S_1}{S_2} = \frac{2(n+2)}{2n+1} \]