product of terms taken $3$ at a time in polynomial expression 
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$ 

Try: 
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdots+98\cdot 99\cdot 100)x^{98}+\cdots$$
for $1$ at a time (Coefficient of $x^{99}$) is $\displaystyle \sum^{100}_{i=1}i = 50\cdot 101$
for $2$ at a time (Coefficient of $x^{98}$) is $\displaystyle \sum^{100}_{i=1}\sum^{100}_{j=1\;, (1\leq i<j \leq 100)}i \cdot j= \frac{1}{2}\bigg[\bigg(\sum^{100}_{i=1}i\bigg)^2-\sum^{100}_{i=1}i^2\bigg]$
But could some help me how to find coefficient of $x^{97},$ thanks
 A: The coefficient of $x^{97}$ is
$$
\begin{align}
\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ijk
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ij\binom{k}{1}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\,((j-2)+2)\binom{j}{2}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\left[3\binom{j}{3}+2\binom{j}{2}\right]\\
&=\sum_{i=1}^{100}i\left[3\binom{i}{4}+2\binom{i}{3}\right]\\
&=\sum_{i=1}^{100}\left[3((i-4)+4)\binom{i}{4}+2((i-3)+3)\binom{i}{3}\right]\\
&=\sum_{i=1}^{100}\left[15\binom{i}{5}+12\binom{i}{4}+8\binom{i}{4}+6\binom{i}{3}\right]\\
&=15\binom{101}{6}+20\binom{101}{5}+6\binom{101}{4}\\[9pt]
&=20618771250
\end{align}
$$
A: With some manipulation (using inclusion exclusion principle) you can write the sum as:
$$\left(\sum_{i=1}^{100} i\right)^3 = \binom{3}{1}\left(\sum_{i=1}^{100} i^2\right)\left(\sum_{i=1}^{100} i\right) - 2\left(\sum_{i=1}^{100}i^3\right) + \binom{3}{1} \binom{2}{1}S $$
Where $S$ is the required sum.
A: If you try to expand the multiplications, you'll find that to produce terms with $x^{98}$, you'll need $2$ distinct constant factors (from the set $\{1,2,\dots,100\})$ and $98$ number of $x$ factors. So the sum of these terms, $a_{98}$, would be $$a_{98}=\sum_{n=1}^{100}\sum_{m=n+1}^{100}mn$$
How can we find this sum? Note that $$(1+2+…100)^2= 1^2+2^2+… 100^2 + 2a_{98}$$ $$\implies a_{98} = \frac12 \left[(5050)^2- \sum_{i=1}^{100} i^2 \right]$$ $$\implies \boxed{ a_{98} = 12582075}$$
