2021학년도 11월 가형 21번

21. 수열 $\{a_n\}$$\{a_n\}$은 $0<a_1<1$$0<a_1<1$이고, 모든 자연수 $n$$n$에 대하여 다음 조건을 만족시킨다.

(가) $a_{2n}=a_2\times a_n+1$$a_{2n}=a_2\times a_n+1$

(나) $a_{2n+1}=a_2\times a_n-2$$a_{2n+1} =a_2\times a_n-2$

$a_8-a_{15}=63$$a_8-a_{15}=63$일 때, $\frac{a_8}{a_1}$$\cfrac {a_8}{a_1}$의 값을 구하시오.

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i. 생각

• 조건에 따라 $a_8,a_{15}$$a_8,~a_{15}$를 정리하자.

  • $a_8$$a_8$을 정리하자.

    $a_8=a_2\times a_4+1$$a_8=a_2\times a_4 +1$

    $a_4=a_2\times a_2+1$$a_4 =a_2\times a_2 +1$

    $a_2=a_2\times a_1+1$$a_2=a_2\times a_1 +1$

  • $a_{15}$$a_{15}$를 정리하자.

    $a_{15}=a_2\times a_7-2$$a_{15}=a_2\times a_7 -2$

    $a_7=a_2\times .a_3-2$$a_7=a_2\times. a_3 -2$

    $a_3=a_2\times a_1-2$$a_3=a_2\times a_1-2$

• $a_8-a_{15}=63$$a_8-a_{15}=63$을 정리하자

  $\begin{array}{rl}{a}_8-a_{15}& =(a_2\times a_4+1)-(a_2\times a_7-2)\\ \\ & =a_2(a_4-a_7)+3\\ \\ & =63\end{array}$$\begin{aligned}a_8-a_{15} &=(a_2\times a_4 +1)-(a_2\times a_7 -2) \\ \\ &= a_2 (a_4 -a_7)+3\\ \\ &=63 \end{aligned}$

  • $a_2(a_4-a_7)=60$$a_2(a_4-a_7)=60$

    $\begin{array}{rl}{a}_4-a_7& =a_2(a_2-a_3)+3\end{array}$$\begin{aligned} a_4-a_7&=a_2(a_2-a_3)+3 \end{aligned}$

    $a_2-a_3=3$$a_2-a_3=3$

    $\therefore a_4-a_7=3a_2+3$$\therefore~ a_4-a_7=3a_2+3$

    $a_2(3a_2+3)=60$$a_2(3a_2+3)=60$

    $\therefore a_2^2+a_2-20=0$$\therefore~ a_2^2 +a_2-20=0$

    $a_2=4,-5$$a_2=4,~-5$

  • $a_2=4$$a_2=4$일 때,

    $a_2=a_2\times a_1+1$$a_2=a_2\times a_1+1$을 이용하면,

    $4=4a_1+1$$4=4a_1+1$

    $a_1=\frac{3}{4}$$a_1=\cfrac 34$

    $a_4=17$$a_4=17$

    $a_8=69$$a_8=69$

    $\therefore \frac{a_8}{a_1}=\frac{69}{\frac{3}{4}}=92$$\therefore~ \cfrac {a_8}{a_1}=\cfrac {69}{\frac 34}=92$

  • $a_2=-5$$a_2=-5$일 때,

    $-5=-5a_1+1$$-5=-5a_1+1$

    $a_1<0$$a_1<0$

    조건에 위배

$\therefore \frac{a_8}{a_1}=92$$\therefore~ \cfrac {a_8}{a_1}=92$