Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$

Question

I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$

Here is how I tried, since for $x=1,2,\ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$ $$f(2)=1-2+\frac{1}{2!}2(2-1)=0$$ $$f(3)=1-3+\frac{1}{2!}3(3-1)-\frac{1}{3!}3(3-2)(3-1)=\binom{3}{0}-\binom{3}{1}+\binom{3}{2}-\binom{3}{3}$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0\,$ and indeed that $f(101)=(1-1)^{101}=0$

Since the polynomial is also of order $101$, the roots are $x=1,2, \ldots, 101$ giving:$$S(f(x))=\sum_{j=1}^{101}j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$$

Answer 1

We have

$$f(x) = a_{101}x^{101} + a_{100}x^{100} + \cdots$$

and we know that $a_{101} = -\frac1{101!}$. Vieta's formulas give $$5151 = \text{ sum of roots of } f = -\frac{a_{100}}{a_{101}}$$

so $a_{100} = \frac{5151}{101!}$.

The derivative is

$$f'(x) = 101a_{101}x^{100} + 100a_{100}x^{99} + \cdots$$

so Vieta's formulas give $$\text{ sum of roots of } f' = -\frac{100a_{100}}{101a_{101}} = -\frac{100\cdot \frac{5151}{101!}}{101\cdot \frac{-1}{101!}} = \frac{515100}{101}= 5100$$

Answer 2

According to Vieta given the polynomial

$$ p_n(x) = a_n x^n+a_{n-1}x^{n-1}+\cdots + a_0\\ p'_n(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+a_1 $$

now

$$ S_{100}(x) = -\frac{a_{100}}{a_{101}}\\ S'_{100}(x) = -\frac{(101-1)a_{100}}{101 a_{101}} $$

Here

$$ a_{101} = \frac{(-1)^{101}}{101!}\\ a_{100} = -a_{101}\frac{100(100+1)}{2}+\frac{(-1)^{100}}{100!} $$

then

$$ S_n(x) = 5151\\ S'_n(x) = 5100 $$

and finally

$$ S_{100}(x)-S'_{100}(x) = 5151-5100 = 51 $$