数学問題22番
問題
計算
n回の試行後、A , B , C , D にいる確率を\(a_{n}\)、\(b_{n}\)、\(c_{n}\)、\(d_{n}\)とおく。
対称性より\(b_{n}=d_{n}\)
漸化式立式します。
① \(a_{n+1}=\displaystyle\frac{4}{5}b_{n}+\displaystyle\frac{1}{5}c_{n}\)
② \(b_{n+1}=\displaystyle\frac{2}{5}a_{n}+\displaystyle\frac{1}{5}b_{n}+\displaystyle\frac{2}{5}c_{n}\)
③ \(a_{n+1}=\displaystyle\frac{4}{5}b_{n}+\displaystyle\frac{1}{5}a_{n}\)
また、\(a_{1}=0\)、\(b_{1}=\displaystyle\frac{2}{5}\)、\(c_{1}=\displaystyle\frac{1}{5}\)
①-③を計算すると、
\(a_{n+1}-c_{n+1}=-\displaystyle\frac{1}{5}(a_{n}-c_{n})\)
よって
\(a_{n}-c_{n}=\biggl(\displaystyle\frac{1}{5}\biggr)^n\) ……④
①から、\(b_{n}=\displaystyle\frac{5}{4}\biggl(a_{n+1}-\displaystyle\frac{1}{5}c_{n}\biggr)\)
これを②の \(b_{n+1}\) と\(b_{n}\) に代入する。
\(\displaystyle\frac{5}{4}a_{n+2}-\displaystyle\frac{1}{4}c_{n+1}=\displaystyle\frac{2}{5}a_{n}+\displaystyle\frac{2}{5}c_{n}+\displaystyle\frac{1}{4}a_{n+1}-\displaystyle\frac{1}{20}c_{n}\)
④を使って\(c_{n}\) を消去すると（\(a_{n}\)を求めたいから）
\(\displaystyle\frac{5}{4}a_{n+2}-\displaystyle\frac{1}{2}a_{n+1}-\displaystyle\frac{3}{4}a_{n}=-\displaystyle\frac{3}{10}\biggl(-\displaystyle\frac{1}{5}\biggr)^n\)
\(a_{n+2}-\displaystyle\frac{2}{5}a_{n+1}-\displaystyle\frac{3}{5}a_{n}=-\displaystyle\frac{6}{25}\biggl(-\displaystyle\frac{1}{5}\biggr)^n\)
となる。これを二通りに式変形する。
⑤ \(a_{n+2}-a_{n+1}=-\displaystyle\frac{3}{5}(a_{n+1}-a_{n})-\displaystyle\frac{6}{25}\biggl(-\displaystyle\frac{1}{5}\biggr)^n\)
⑥ \(a_{n+2}+\displaystyle\frac{3}{5}a_{n+1}=(a_{n+1}+\displaystyle\frac{3}{5}a_{n})-\displaystyle\frac{6}{25}\biggl(-\displaystyle\frac{1}{5}\biggr)^n\)
⑤に関して。
\(a_{n+1}-a_{n}=b_{n}\) とおき、両辺に\((-5)^{n+1}\)をかける。
\((-5)^{n+1}b_{n+1}=3(-5)^{n}b_{n}+\displaystyle\frac{6}{5}\)
\((-5)^{n}b_{n}=c_{n}\) とおく。
\(c_{n+1}+\displaystyle\frac{3}{5}=3\biggl(c_{n}+\displaystyle\frac{3}{5}\biggr)\)
\(c_{1}=-5b_{1}=-5(a_{2}-a_{1})=-\displaystyle\frac{9}{5}\) より
\(c_{n}=-\displaystyle\frac{3}{5}-\displaystyle\frac{2}{5}\times 3^n\)
\((-5)^{n}b_{n}=c_{n}\)より
\(b_{n}=-\displaystyle\frac{3}{5}\biggl(-\displaystyle\frac{1}{5}\biggr)^n-\displaystyle\frac{2}{5}\biggl(-\displaystyle\frac{3}{5}\biggr)^n\)
よって
\(a_{n+1}-a_{n}=-\displaystyle\frac{3}{5}\biggl(-\displaystyle\frac{1}{5}\biggr)^n-\displaystyle\frac{2}{5}\biggl(-\displaystyle\frac{3}{5}\biggr)^n\) ……⑦
⑥も同様に計算していく。
\(d_{n}=a_{n+1}+\displaystyle\frac{3}{5}a_{n}\) とおく。
\((-5)^{n+1}d_{n+1}=(-5)(-5)^{n}d_{n}+\displaystyle\frac{6}{5}\)
\((-5)^{n}d_{n}=e_{n}\) とおく。
\(e_{n+1}=-5e_{n}+\displaystyle\frac{6}{5}\)
\(e_{n}=\displaystyle\frac{1}{5}-2(-5)^{n-1}\)
戻すと
\(d_{n}=-\biggl(-\displaystyle\frac{1}{5}\biggr)^{n+1}+\displaystyle\frac{2}{5}\)
よって
\(a_{n+1}+\displaystyle\frac{3}{5}a_{n}=-\biggl(-\displaystyle\frac{1}{5}\biggr)^{n+1}+\displaystyle\frac{2}{5}\) ……⑧
⑦－⑧より
\(\displaystyle\frac{8}{5}a_{n}=\displaystyle\frac{2}{5}+\displaystyle\frac{4}{5}\biggl(-\displaystyle\frac{1}{5}\biggr)^n+\displaystyle\frac{2}{5}\biggl(-\displaystyle\frac{3}{5}\biggr)^n\)
答え
\(a_{n}=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{2}\biggl(-\displaystyle\frac{1}{5}\biggr)^n+\displaystyle\frac{1}{4}\biggl(-\displaystyle\frac{3}{5}\biggr)^n\)
