Determine the sum of the roots of any f'

Question

Knowing that we have the polynomial $f(x)=a(x-1)(x-2)\dots(x-101)$ with $1,2,\dots101$ it's roots , how do I determine the sum of the roots of $f'$? Yesterday on one of my posts I got an idea of explication but I didn't understand it . Is there a specific formula? Here you can find in the comments the answer , but I don't know how he got to it What would be the roots of the derivative of this polynomial ... is the one with the sum equals to 5100

Answer 1

Hint:

Observe that $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc.$$

So if you know the roots of a cubic polynomial, their sum is the opposite of the quadratic coefficient. This rule generalizes to polynomials of any degree. If the leading coefficient is not unit, divide by it to normalize.

Now you know that $f$ is (ignoring $a$)

$$x^{101}-5151\,x^{100}+\cdots$$

frow wich you determine $f'$.

Answer 2

The formula is simple

The average of the roots of a polynomial of degree $n>1$ is the same as the average of the roots of its derivative.

In the question on hand the average of roots of the polynomial is $51$

Therefore the average of roots of its derivative is also $51$

Since the derivative has only $100$ roots, the sum of roots of the derivative is $5100$