Coeff. of $x^{97}$ in $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)$

Solution 1:

For the coefficient of $x^{97}$ you can apply a method similar to the one you used to find the coefficient of $x^{98}$. You already noted that the coefficient $C$ of $x^{97}$ is $$C:=-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k=-\sum_{i=1}^{98}\sum_{j=i+1}^{99}\sum_{k=j+1}^{100}i\cdot j\cdot k.$$ We can compute $C$ from the identity $$\left(\sum_{i=1}^{100}i\right)^3=-2\sum_{i=1}^{100}i^3+3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)-6C.$$ Rewriting the above yields \begin{eqnarray*} C&=&-\frac{1}{6}\left[\left(\sum_{i=1}^{100}i\right)^3+2\sum_{i=1}^{100}i^3-3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)\right],\\ &=&-\frac{1}{6}\left[\left(\tfrac{1}{2}\cdot100\cdot101\right)^3+2\cdot\left(\tfrac{1}{4}\cdot100^2\cdot101^2\right)-3\cdot\left(\tfrac{1}{6}\cdot100\cdot101\cdot201\right)\cdot\left(\tfrac{1}{2}\cdot100\cdot101\right)\right]\\ &=&-20618771250, \end{eqnarray*} modulo miscalculations.