Question

Asta e întrebarea cu polinoame​

Answer 1

$\displaystyle f(x)= x^n-2x+2 \\ \\ S = \sum\limits_{k=1}^n\dfrac{1}{(1-x_k)^2}\\ \\\\f(x) = (x-x_1)(x-x_2)(x-x_3)\cdot...\cdot(x-x_n) \\ \\ f'(x) = \sum\limits_{k=1}^{n}\dfrac{f(x)}{x-x_k}\\ \\ f''(x) = \sum\limits_{k=1}^{n}\dfrac{f'(x)(x-x_k) - f(x)}{(x-x_k)^2} \\ \\ f''(x) = \sum\limits_{k=1}^{n}\dfrac{f'(x)}{x-x_k}-\sum\limits_{k=1}^{n}\dfrac{f(x)}{(x-x_k)^2}\\ \\ \\\text{Facem }x = 1:$

$\displaystyle f''(1) = \sum\limits_{k=1}^{n}\dfrac{f'(1)}{1-x_k}-\sum\limits_{k=1}^{n}\dfrac{f(1)}{(1-x_k)^2}\\ \\ n(n-1) = \sum\limits_{k=1}^{n}\dfrac{n-2}{1-x_k} - \sum\limits_{k=1}^n\dfrac{1}{(1-x_k)^2} \\ \\ n(n-1)=(n-2)\cdot \sum\limits_{k=1}^n\dfrac{1}{1-x_k} - S\\ \\ n(n-1)=(n-2)\cdot \sum\limits_{k=1}^n\dfrac{f(1)}{1-x_k} - S \\ \\ n(n-1)=(n-2)\cdot f'(1) - S\\ \\ n(n-1)=(n-2)\cdot (n-2) - S\\ \\ n^2-n = n^2-4n+4 - S \\ \\ \Rightarrow \boxed{S = 4-3n}$