Si $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)\;,$, Entonces el Coeficiente de $x^{99}$ y
Coeficiente de $x^{98}$ y el Coeficiente de $x^{97}$ $f(x).$
$\bf{My\; try::}$ Podemos escribir $f(x)$
$$\displaystyle f(x) = x^{100}-\left(\sum_{i=1}^{100}i\right)x^{99}+\left(\mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j\right)x^{98}-\left(\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k\right)x^{97}+..$$
Por lo $\bf{Coefficients}$ $$\displaystyle x^{99} = -\sum_{i=1}^{100}\left(1+2+3+......+100\right) = -\frac{100\cdot 101}{2} = -5050$$
Del Mismo Modo Coef. de $$\displaystyle x^{98} = \mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j = \frac{1}{2}\left[\left(\sum_{i=1}^{100}i\right)^2-\sum_{i=1}^{100}i^2\right] = \frac{1}{2}\left[\left(\frac{100\cdot 101}{2}\right)^2-\frac{100\cdot 101\cdot 201}{6}\right]$$
Pero no entiendo Cómo puedo calcular Coef. de $\displaystyle x^{97} =-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k$
Me ayude, Gracias