자연계열 (오후) 2번

(2-1) $\displaystyle P_5(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$에 대해 다음을 얻는다.$\displaystyle Q_5(x)=\frac1{P_5(x)}=\frac{b_1}{x-1}+\frac{b_2}{x-2}+\frac{b_3}{x-3}+\frac{b_4}{x-4}+\frac{b_5}{x-5}$$\displaystyle \Rightarrow\quad1=b_1\frac{P_5(x)}{x-1}+b_2\frac{P_5(x)}{x-2}+b_3\frac{P_5(x)}{x-3}+b_4\frac{P_5(x)}{x-4}+b_5\frac{P_5(x)}{x-5}$$\displaystyle \begin{aligned}\Rightarrow\quad1&=b_1(x-2)(x-3)(x-4)(x-5)\\&\quad+b_2(x-1)(x-3)(x-4)(x-5)\\&\qquad+b_3(x-1)(x-2)(x-4)(x-5)\\&\qquad\quad+b_4(x-1)(x-2)(x-3)(x-5)\\&\qquad\qquad+b_5(x-1)(x-2)(x-3)(x-4)\end{aligned}$$\displaystyle x=1$을 대입하면 $\displaystyle 1=b_1(-1)(-2)(-3)(-4)=(4!)b_1\quad\Rightarrow\quad b_1=\frac1{4!}=\frac1{24}$$\displaystyle x=2$를 대입하면 $\displaystyle 1=b_2(1)(-1)(-2)(-3)=-(3!)b_2\quad\Rightarrow\quad b_2=-\frac1{3!}=-\frac16$$\displaystyle x=3$을 대입하면 $\displaystyle 1=b_3(2)(1)(-1)(-2)=(2!)(2!)b_3\quad\Rightarrow\quad b_3=\frac1{(2!)(2!)}=\frac14$$\displaystyle x=4$를 대입하면 $\displaystyle 1=b_4(3)(2)(1)(-1)=-(3!)b_4\quad\Rightarrow\quad b_4=-\frac1{3!}=-\frac16$$\displaystyle x=5$를 대입하면 $\displaystyle 1=b_5(4)(3)(2)(1)=(4!)b_5\quad\Rightarrow\quad b_5=\frac1{4!}=\frac1{24}$

(2-2)$\displaystyle Q_n(x)=\frac1{P_n(x)}=\frac{b_1}{x-1}+\frac{b_2}{x-2}+\frac{b_3}{x-3}+\cdots+\frac{b_n}{x-n}$$\displaystyle \Rightarrow\quad1=b_1\frac{P_n(x)}{x-1}+b_2\frac{P_n(x)}{x-2}+b_3\frac{P_n(x)}{x-3}+\cdots+b_n\frac{P_n(x)}{x-n}$$\displaystyle \begin{aligned}\Rightarrow\quad1&=b_1(x-2)(x-3)(x-4)\cdots(x-n)\\&\quad+b_2(x-1)(x-3)(x-4)\cdots(x-n)\\&\qquad+b_3(x-1)(x-2)(x-4)\cdots(x-n)\\&\qquad\qquad\vdots\\&\qquad\quad+b_n(x-1)(x-2)(x-3)\cdots(x-(n-1))\end{aligned}$∴ $\displaystyle x=i\ (i=1,2,3,\cdots,n)$이라 놓으면$\displaystyle 1=b_1 0+b_2 0+\cdots+b_{i-1}0+b_i(-1)^{n-i}(i-1)!(n-i)!+b_{i+1}0+\cdots+b_n0$$\displaystyle \Rightarrow\quad b_i=\frac{(-1)^{n-i}}{(i-1)!(n-i)!}=\frac{(-1)^{n-i}}{(n-1)!}\frac{(n-1)!}{(i-1)!(n-i)!}=\frac{(-1)^{n-i}}{(n-1)!}{}_{n-1}\mathrm C_{i-1}$$\displaystyle \Rightarrow\quad(n-1)!\sum_{i=1}^n|b_i|=(n-1)!\sum_{i=1}^n\left[\frac1{(n-1)!}{}_{n-1}\mathrm C_{i-1}\right]=\sum_{i=1}^n{}_{n-1}\mathrm C_{i-1}=2^{n-1}$