자연계열 2번

[문제2-1]먼저, $\displaystyle \frac1{a_1a_2a_3a_4}$의 경우를 살펴보면 아래와 같다.$\displaystyle \begin{aligned}\frac1{a_1a_2a_3a_4}&=\frac1{a_4-a_1}\left(\frac1{a_1a_2a_3}-\frac1{a_2a_3a_4}\right)=\frac1{(a_4-a_1)(a_3-a_1)(a_2-a_1)}\left(\frac1{a_1}-\frac1{a_2}\right)\\&\quad-\left\{\frac1{(a_4-a_1)(a_3-a_1)(a_3-a_2)}+\frac1{(a_4-a_1)(a_4-a_2)(a_3-a_2)}\right\}\left(\frac1{a_2}-\frac1{a_3}\right)\\&\quad+\frac1{(a_4-a_1)(a_4-a_2)(a_4-a_3)}\left(\frac1{a_3}-\frac1{a_4}\right)\end{aligned}$

같은 방법으로 $\displaystyle \frac1{a_1a_2a_3a_4a_5}$의 경우를 살펴보면 아래와 같다.$\displaystyle \begin{aligned}\frac1{a_1a_2a_3a_4a_5}&=\frac1{a_5-a_1}\left(\frac1{a_1a_2a_3a_4}-\frac1{a_2a_3a_4a_5}\right)\\&=\frac1{(a_5-a_1)(a_4-a_1)(a_3-a_1)(a_2-a_1)}\left(\frac1{a_1}-\frac1{a_2}\right)\\&\quad-\left\{\frac1{(a_5-a_1)(a_4-a_1)(a_3-a_1)(a_3-a_2)}+\frac1{(a_5-a_1)(a_4-a_1)(a_4-a_2)(a_3-a_2)}\right.\\&\left.\quad\qquad+\frac1{(a_5-a_1)(a_5-a_2)(a_4-a_2)(a_3-a_2)}\right\}\left(\frac1{a_2}-\frac1{a_3}\right)\\&\quad+\left\{\frac1{(a_5-a_1)(a_4-a_1)(a_4-a_2)(a_4-a_3)}+\frac1{(a_5-a_1)(a_5-a_2)(a_4-a_2)(a_4-a_3)}\right.\\&\left.\quad\qquad+\frac1{(a_5-a_1)(a_5-a_2)(a_5-a_3)(a_4-a_3)}\right\}\left(\frac1{a_3}-\frac1{a_4}\right)\\&\quad-\frac1{(a_5-a_1)(a_5-a_2)(a_5-a_3)(a_5-a_4)}\left(\frac1{a_4}-\frac1{a_5}\right)\end{aligned}$

따라서, $\displaystyle X_1$, $\displaystyle X_2$, $\displaystyle X_3$, $\displaystyle X_4$는 아래와 같다.$\displaystyle X_1=\frac1{(a_5-a_1)(a_4-a_1)(a_3-a_1)(a_2-a_1)}$$\displaystyle \begin{aligned}X_2&=-\frac1{(a_5-a_1)(a_4-a_1)(a_3-a_1)(a_3-a_2)}-\frac1{(a_5-a_1)(a_4-a_1)(a_4-a_2)(a_3-a_2)}\\&\quad-\frac1{(a_5-a_1)(a_5-a_2)(a_4-a_2)(a_3-a_2)}\end{aligned}$$\displaystyle \begin{aligned}X_3&=\frac1{(a_5-a_1)(a_4-a_1)(a_4-a_2)(a_4-a_3)}+\frac1{(a_5-a_1)(a_5-a_2)(a_4-a_2)(a_4-a_3)}\\&\quad+\frac1{(a_5-a_1)(a_5-a_2)(a_5-a_3)(a_4-a_3)}\end{aligned}$$\displaystyle X_4=-\frac1{(a_5-a_1)(a_5-a_2)(a_5-a_3)(a_5-a_4)}$

[문제2-2]위 계산식에 $\displaystyle a_i=x-i$을 대입하여 정리하면 다음이 성립한다.$\displaystyle \frac1{a_1a_2a_3a_4}=-\frac1{3!}\left(\frac1{x-1}-\frac3{x-2}+\frac3{x-3}-\frac1{x-4}\right)$$\displaystyle \frac1{a_1a_2a_3a_4a_5}=\frac1{4!}\left(\frac1{x-1}-\frac4{x-2}+\frac6{x-3}-\frac4{x-4}+\frac1{x-5}\right)$따라서 다음이 성립함을 추측할 수 있고, 이를 수학적 귀납법을 사용하여 증명할 수 있다.$\displaystyle \frac1{a_1a_2\cdots a_n}=\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n-1}\mathrm C_k\frac1{x-k-1}$ (단, $\displaystyle n$은 $\displaystyle 2$이상의 자연수)($\displaystyle {}_{0}\mathrm C_0=1$으로 정하면 $\displaystyle n=1$일 때도 상기 식은 $\displaystyle \frac1{a_1}=\frac1{x-1}$로 성립함.)

이로부터 다음 식을 얻는다.$\displaystyle \begin{aligned}\int_{n+1}^{n+2}\frac1{a_1a_2\cdots a_n}\,dx&=\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n-1}\mathrm C_k\int_{n+1}^{n+2}\frac1{x-k-1}\,dx\\&=\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n-1}\mathrm C_k(\ln(n-k+1)-\ln(n-k))\end{aligned}$이때 $\displaystyle {}_{n-1}\mathrm C_{k-1}+{}_{n-1}\mathrm C_k={}_{n}\mathrm C_k$임을 이용하면$\displaystyle =\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n}\mathrm C_k\ln(n-k+1)$$\displaystyle =\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n-1}\mathrm C_k(\ln(n-k+1)-\ln(n-k))$이때 $\displaystyle {}_{n-1}\mathrm C_{k-1}+{}_{n-1}\mathrm C_k={}_{n}\mathrm C_k$임을 이용하면$\displaystyle =\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k{}_{n}\mathrm C_k\ln(n-k+1)$