Sum of a set of products

$\large \displaystyle S = \sum_{a=1}^{n} \sum_{b=1}^{n} a \cdot b - \sum_{a=1, a=b}^{n} \sum_{b=1, b=a}^{n} a \cdot b$

$\large \displaystyle S = \sum_{a=1}^{n} a \sum_{b=1}^{n} b - \sum_{a=1}^{n} a^2$

$\large \displaystyle S = \frac{n \cdot (n+1)}{2} \cdot \frac{n \cdot(n+1)}{2} - \frac{ n \cdot(n+1) \cdot (2n+1) }{6}$

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{(n^3-n)(3n+2)}{12}}$

For $n=99$ :

$\color{#20A900} \boxed{ \large \displaystyle S = 24174150 }$

Thus:

$\color{#3D99F6} \large \displaystyle m = 2417.415, \boxed{ \large \displaystyle \lfloor m \rfloor = 2417}$

We have that $a, b \in \{ \mathbb{Z^+} < n \}$ and $a \neq b$ and we want to find the sum of the product of each pair of integers $(a,b)$ . In other words, we want to find $\large \displaystyle \sum_{\substack{a,b \space \in \space \{ \mathbb{Z^+} < \space n \} \\ a \neq b}}{a \cdot b}$ .

Define a set $S$ such that $S_i \in \{ \mathbb{Z^+} < n \}$ , where $i$ refers to the $i^{\text{th}}$ element (the index) of $S$ .

The sum of the product of each pair is thus $\displaystyle \sum_{\substack{a,b \space \in \space \{ \mathbb{Z^+} < \space n \} \\ a \neq b}}{a \cdot b} = \sum_{\substack{a,b \space \in \space S \\ a \neq b}}{a \cdot b} = \sum_{i \neq j}{S_i \cdot S_j}$ .

Note that the square of the sum of all the elements of $S$ (positive integers less than $n$ ) is $\displaystyle \sum_{j = 1}^{n-1}{j^2} + \sum_{i \neq j}{S_i \cdot S_j}$ .

So $\displaystyle \sum_{i \neq j}{S_i \cdot S_j} = \left( \sum_{j = 1}^{n-1}{j} \right)^2 - \sum_{j = 1}^{n-1}{j^2}$ .

From here, we can reduce $\displaystyle \sum_{i \neq j}{S_i \cdot S_j}$ to $\left( \dfrac{n(n-1)}{2} \right) ^2 - \dfrac{n(n-1)(2n-1)}{6}$ .

Therefore $\displaystyle \sum_{i \neq j}{S_i \cdot S_j} = \dfrac{n(n-1)\left(3n(n-1)-2(2n-1)\right)}{12} = \dfrac{(n-2)(n-1)n(3n-1)}{12}$ (Final term as pointed out by @Mark Hennings ).

For $n = 100$ this is $\dfrac{98 \cdot 99 \cdot 100 \cdot 299}{12} = 33 \cdot 49 \cdot 50 \cdot 299$ and this value is $m \times 10^4$ .

$m$ is thus $\dfrac{49 \cdot 299 \cdot 1650}{10^4} = \dfrac{14651 \cdot 165}{1000}$ .

By direct computation, we see that $\lfloor m \rfloor = 2417._\square$