Răspuns:
Fie [tex]f(x)=x^n-3x^{n-1}+2x+1[/tex] și [tex]x_1, \ x_2, \ \ldots, x_n[/tex] rădăcinile sale. Atunci
[tex]f(x)=\left(x-x_1\right)\left(x-x_2\right)\ldots\left(x-x_n\right)[/tex]
[tex]f'(x)=(x-x_2)(x-x_3)\ldots(x-x_n)+(x-x_1)(x-x_3)\ldots(x-x_n)+\ldots+(x-x_1)(x-x_2)\ldots(x-x_{n-1})[/tex]
Atunci
[tex]\displaystyle\frac{f'(x)}{f(x)}=\displaystyle\frac{1}{x-x_1}+\frac{1}{x-x_2}+\ldots\frac{1}{x-x_n}[/tex]
Avem
[tex]S=\displaystyle\sum_{k=1}^n\frac{x_k}{x_k-1}=\sum_{k=1}^n\frac{x_k-1+1}{x_k-1}=\sum_{k=1}^n\left(1+\frac{1}{x_k-1}\right)=n-\sum_{k=1}^n\frac{1}{1-x_k}}[/tex]
Dar
[tex]\displaystyle\sum_{k=1}^n\frac{1}{1-x_k}=\frac{f'(1)}{f(1)}=-2n+5[/tex]
Rezultă [tex]S=3n-5[/tex].
Explicație pas cu pas:
