2021年入試問題研究に戻る

東北大数学AO 4番解答

(1)　 $n\geqq 1$のとき，部分積分より， \begin{eqnarray*} K_n&=&\int_0^{\frac{\pi}{2}} x^n(-\cos x)'\,dx =\left[-x^n\cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} nx^{n-1}\cos x\,dx\\ &=&n\int_0^{\frac{\pi}{2}} x^{n-1}\cos x\,dx\\ &=&nL_{n-1}\\ L_n&=&\int_0^{\frac{\pi}{2}} x^n(\sin x)'\,dx =\left[x^n\sin x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} nx^{n-1}\sin x\,dx\\ &=&\left(\dfrac{\pi}{2} \right)^n-n\int_0^{\frac{\pi}{2}} x^{n-1}\sin x\,dx\\ &=&\left(\dfrac{\pi}{2} \right)^n-nK_{n-1} \end{eqnarray*} よって \begin{eqnarray*} K_{n+1}&=&(n+1)L_n=(n+1)\left(\dfrac{\pi}{2} \right)^n-n(n+1)K_{n-1}\\ L_{n+1}&=&\left(\dfrac{\pi}{2} \right)^{n+1}-(n+1)K_n= \left(\dfrac{\pi}{2} \right)^{n+1}-n(n+1)L_{n-1} \end{eqnarray*}

(2)　 定義より， \begin{eqnarray*} K_0&=&\int_0^{\frac{\pi}{2}} \sin x\,dx=\left[-\cos x \right]_0^{\frac{\pi}{2}}=1\\ L_0&=&\int_0^{\frac{\pi}{2}} \cos x\,dx =\left[\sin x \right]_0^{\frac{\pi}{2}}=1 \end{eqnarray*} である．また(1)から \begin{eqnarray*} K_1&=&L_0=1\\ L_1&=&\dfrac{\pi}{2}-K_0=\dfrac{\pi}{2}-1 \end{eqnarray*} である．(1)の結果より， \[ \dfrac{K_{n+1}}{(n+1)!}=\dfrac{1}{n!}\left(\dfrac{\pi}{2} \right)^n-\dfrac{K_{n-1}}{(n-1)!} \] これから，

i)　$n$が偶数のとき \begin{eqnarray*} \dfrac{K_n}{n!} &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{K_{n-2}}{(n-2)!}\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}+(-1)^{\frac{3-1}{2}}\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+\dfrac{K_{n-4}}{(n-4)!}\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n-1-1}{2}}\left(\dfrac{\pi}{2}-K_0 \right)\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n}{2}-1}\cdot \left(\dfrac{\pi}{2}-1\right) \end{eqnarray*}

ii)　$n$が奇数のとき，同様に \begin{eqnarray*} \dfrac{K_n}{n!} &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{K_{n-2}}{(n-2)!}\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}+(-1)^{\frac{3-1}{2}}\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+\dfrac{K_{n-4}}{(n-4)!}\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n-1}{2}}K_1\\ &=&\dfrac{1}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{1}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n-1}{2}} \end{eqnarray*} よって， \[ K_n=\left\{ \begin{array}{ll} \dfrac{n!}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{n!}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n}{2}-1}\cdot n!\left(\dfrac{\pi}{2}-1\right)&(n:偶数)\\ \dfrac{n!}{(n-1)!}\left(\dfrac{\pi}{2} \right)^{n-1}-\dfrac{n!}{(n-3)!}\left(\dfrac{\pi}{2} \right)^{n-3}+ \cdots +(-1)^{\frac{n-1}{2}}n!&(n:奇数) \end{array} \right. \] となる．

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