2021학년도 09월 나형 21번

21. 수열 $\{a_n\}$$\{a_n\}$은 모든 자연수 $n$$n$에 대하여

$$\begin{array}{c}{a}_{n+2}=\{\begin{array}{ll}2a_n+a_{n+1}& (a_n\le a_{n+1})\\ \\ a_n+a_{n+1}& (a_n>a_{n+1})\end{array}\end{array}$$

을 만족시킨다. $a_3=2,a_6=19$$a_3=2,~a_6=19$가 되도록 하는 모든 $a_1$$a_1$의 값을 구하시오.

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i. 생각

$a_1,a_2,a_3=2,a_4,a_5,a_6=19$$a_1,~a_2,~a_3=2,~a_4,~a_5,~a_6=19$

아..........경우를 상당히 나눠야 하나...

• $a_1\le a_2$$a_1\le a_2$

  $2a_1+a_2=a_3=2$$2a_1+a_2=a_3=2$

  $a_2=2-2a_1$$a_2 =2-2a_1$

  • $a_2$$a_2$와 $a_3$$a_3$의 대소비교

    • $a_1<0$$a_1 <0$

      • $a_2>a_3$$a_2>a_3$

        $a_2+a_3=a_4$$a_2+a_3=a_4$

        $a_4=4-2a_1$$a_4=4-2a_1$

      • $a_3<a_4$$a_3<a_4$

        $a_5=8-2a_1$$a_5=8-2a_1$

      • $a_4<a_5$$a_4<a_5$

        $a_6=16-6a_1=19$$a_6 =16-6a_1=19$

      $\therefore a_1=-\frac{1}{2}$$\therefore~ a_1=-\cfrac 12$

    • $a_1\ge 0$$a_1\ge 0$

      $a_2=2-2a_1$$a_2=2-2a_1$

      • $a_2<a_3$$a_2<a_3$

        $a_4=6-4a_1$$a_4=6-4a_1$

      • $a_3$$a_3$과 $a_4$$a_4$의 대소비교

        $\begin{array}{rl}{a}_4-a_3& =4-4a_1\end{array}$$\begin{aligned}a_4-a_3&=4-4a_1 \end{aligned}$

        • $0\le a_1\le 1$$0\le a_1 \le 1$

          • $a_3<a_4$$a_3<a_4$

            $a_5=10-4a_1$$a_5=10-4a_1$

            $a_4<a_5$$a_4<a_5$

            $\therefore a_6=22-12a_1=19$$\therefore~ a_6=22-12a_1=19$

          $\therefore a_1=\frac{1}{4}$$\therefore~ a_1=\cfrac 14$

        • $1<a_1$$1<a_1$

          • $a_3>a_4$$a_3>a_4$

            $a_5=8-4a_1$$a_5=8-4a_1$

            • $a_4<a_5$$a_4<a_5$

              $a_6=20-12a_1=19$$a_6=20-12a_1=19$

          $\therefore a_1=\frac{1}{12}$$\therefore~ a_1=\cfrac 1{12}$

          그런데 모순

• $a_1>a_2$$a_1>a_2$일 때,

  $a_2=2-a_1$$a_2=2-a_1$

  • $a_1>1$$a_1>1$일 때,

    $a_2<2=a_3$$a_2<2=a_3$

    $a_4=6-2a_1$$a_4=6-2a_1$

    • $a_3$$a_3$과 $a_4$$a_4$의 대소비교

      $6-2a_1-2=4-2a_1$$6-2a_1-2=4-2a_1$

      • $1<a_1\le 2$$1<a_1 \le 2$일 때,

        • $a_4\ge a_3$$a_4\ge a_3$

          $a_5=10-2a_1$$a_5=10-2a_1$

          • $a_5>a_4$$a_5>a_4$

            $a_6=22-6a_1=19$$a_6=22-6a_1=19$

        $\therefore a_1=\frac{1}{2}$$\therefore~ a_1=\cfrac 12$

        그런데 모순

      • $2<a_1$$2<a_1$

        • $a_3>a_4$$a_3>a_4$

          $a_5=8-2a_1$$a_5=8-2a_1$

          • $a_4<a_5$$a_4<a_5$

            $a_6=16-6a_1=19$$a_6=16-6a_1=19$

        $\therefore a_1=-\frac{1}{2}$$\therefore~ a_1=-\cfrac 12$

        모순

$\therefore a_1=-\frac{1}{2},\frac{1}{4}$$\therefore~ a_1=-\cfrac 12,~\cfrac 14$

$\therefore -\frac{1}{4}$$\therefore~ -\cfrac 14$