Can we stop with polynomials!

If you don't know any formulas you can solve this in the following way:

First expand $\left(x-1 \right)*(x-2)*(x-3)$ ... to find out that this becomes: $x^{3}-6x^{2}+...$ In this case, the coefficient of the second highest power of $x$ (that is, $x^{2}$ ) is $-1-2-3=-6$ Therefore we can assume that the value of the coefficient of the second highest power of $x$ (wich I will call $A$ ) in a polynomial expansion with the form $\left(x-a_{1}\right)*(x-a_{2})*...*(x-a_{n}) = x^{n}-Ax^{n-1}+...$ is simply the sum of all the terms not dependent on $x$ in the parenthesis: $A=-a_{1}-a_{2}-...-a_{n}$ In the case of this problem, the terms $a_{1}, a_{2}, a_{3}...a_{100}$ are equal to $-1, -2, -3,...,-100$ . This is a arithmetic series with $d=-1$ and $a_{1} = -1$ .

Using the formula to calculate this arithmetic series results in: $A=\frac{-n* \left(a_{1}+a_{n} \right)}{2} =\frac{-100*\left(-1-100\right)}{2}$ $\boxed{A=-5050}$

We notice that when this polynomial is written in expanded form, we get: $x^{100}+bx^{99}+...$ , where b is the coefficient of $x^{99}$ . Vieta's theorem states that the sum of the roots of this polynomial is equal to $\frac{-b}{1}$ . Because of the expanded form the problem was originally given in, we quickly see that the sum of roots is given by $1+2+3+...+100=\frac{100*(100+1)}{2}=5050$ . Thus $-b=5050$ , and $\boxed{b = -5050}$

By Vieta's formula, we know that $\displaystyle a_{99} = -\sum_{i=1}^{100} x_{i} = -\sum_{n=1}^{100} n = -\frac{100(101)}{2} = \boxed{-5050}$ , where $x_{i}$ are the roots (we can ignore $a_{100}$ since it is a monic polynomial).