If `S_(n)` denotes the sum of the products of the products of the first n natureal number taken two at a time

Options

A. `(11e)/(24)`
B. `(11e)/(12)`
C. `(13e)/(24)`
D. none of these

Answer

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

### Step 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 \( S_n = \sum_{1 \leq i < j \leq n} ij \).

### Step 2: Using the Identity
We can use the identity:
\[ (a + b + c + \ldots)^2 = a^2 + b^2 + c^2 + \ldots + 2(ab + ac + bc + \ldots) \]
In our case, let \( a = 1, b = 2, c = 3, \ldots, n \). Therefore, we can express the square of the sum of the first \( n \) natural numbers as:
\[ (1 + 2 + 3 + \ldots + n)^2 = \sum_{k=1}^{n} k^2 + 2S_n \]

### Step 3: Calculate the Sum of the First \( n \) Natural Numbers
The sum of the first \( n \) natural numbers is given by:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus,
\[ \left(\frac{n(n + 1)}{2}\right)^2 = \frac{n^2(n + 1)^2}{4} \]

### Step 4: Calculate the Sum of the Squares of the First \( n \) Natural Numbers
The sum of the squares of the first \( n \) natural numbers is given by:
\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Set Up the Equation
Substituting these results into our identity gives:
\[ \frac{n^2(n + 1)^2}{4} = \frac{n(n + 1)(2n + 1)}{6} + 2S_n \]

### Step 6: Rearranging the Equation
Rearranging the equation to isolate \( S_n \):
\[ 2S_n = \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{6} \]

### Step 7: Finding a Common Denominator
The common denominator for the fractions on the right side is 12:
\[ 2S_n = \frac{3n^2(n + 1)^2}{12} - \frac{2n(n + 1)(2n + 1)}{12} \]
Combining these fractions gives:
\[ 2S_n = \frac{3n^2(n + 1)^2 - 2n(n + 1)(2n + 1)}{12} \]

### Step 8: Simplifying the Numerator
Now we simplify the numerator:
\[ 3n^2(n + 1)^2 - 2n(n + 1)(2n + 1) = 3n^2(n^2 + 2n + 1) - 2n(2n^2 + 3n + 1) \]
Expanding both terms:
\[ = 3n^4 + 6n^3 + 3n^2 - (4n^3 + 6n^2 + 2n) = 3n^4 + 2n^3 - n^2 - 2n \]

### Step 9: Final Expression for \( S_n \)
Thus, we have:
\[ S_n = \frac{3n^4 + 2n^3 - n^2 - 2n}{24} \]

### Step 10: Conclusion
The final expression for the sum \( S_n \) of the products of the first \( n \) natural numbers taken two at a time is:
\[ S_n = \frac{n(n + 1)(n - 1)(3n + 2)}{24} \]