2020년 04월 가형 30번

30. 두 수열 $\{a_n\},\{b_n\}$$\{a_n\},~\{b_n\}$이 모든 자연수 $n$$n$에 대하여 다음 조건을 만족시킨다.

(가) $a_{2n}=b_n+2$$a_{2n}=b_n+2$

(나) $a_{2n+1}=b_n-1$$a_{2n+1} =b_n-1$

(다) $b_{2n}=3a_n-2$$b_{2n}=3a_n-2$

(라) $b_{2n+1}=-a_n+3$$b_{2n+1}=-a_n+3$

$a_{48}=9$$a_{48}=9$이고, $\sum _{n=1}^{63}{a}_n-\sum _{n=1}^{31}{b}_n=155$$\sum\limits_{ n=1}^{63}a_n -\sum\limits_{ n=1}^{31}b_n =155$일 때, $b_{32}$$b_{32}$의 값을 구하시오.

---

i. 생각

• 우선 $b_{32}$$b_{32}$의 값을 구하기 위해 필요한 값을 찾도록 하자.

  $b_{32}=3a_{16}-2$$b_{32}=3a_{16}-2$

  $a_{16}=b_8+2$$a_{16}=b_8+2$

  $b_8=3a_4-2$$b_8=3a_4-2$

  $a_4=b_2+2$$a_4=b_2+2$

  $b_2=3a_1-2$$b_2=3a_1-2$

  $a_1$$a_1$의 값이 필요하다.

• $a_{48}$$a_{48}$도 이용하자.

  $a_{48}=b_{24}+2=9⟶b_{24}=7$$a_{48}=b_{24}+2=9\quad \longrightarrow \quad b_{24}=7$

  $b_{24}=3a_{12}-2=7⟶a_{12}=3$$b_{24}=3a_{12}-2=7\quad \longrightarrow \quad a_{12}=3$

  $a_{12}=b_6+2=3⟶b_6=1$$a_{12}=b_6+2=3\quad \longrightarrow \quad b_6=1$

  $b_6=3a_3-2=1⟶a_3=1$$b_6=3a_3-2=1\quad \longrightarrow \quad a_3=1$

  $a_3=b_1-1=1⟶b_1=2$$a_3=b_1-1=1\quad \longrightarrow \quad b_1=2$

• $\sum _{n=1}^{63}{a}_n-\sum _{n=1}^{31}{b}_n=155$$\sum\limits_{ n=1}^{63}a_n-\sum\limits_{ n=1}^{31}b_n =155$를 이용하자.

  • $\sum _{n=1}^{63}{a}_n$$\sum\limits_{ n=1}^{63}a_n$을 생각하자.

    오호라?

    $a_{2n}+a_{2n+1}=2b_n+1$$a_{2n}+a_{2n+1}=2b_n+1$

    $b_{2n}+b_{2n+1}=2a_n+1$$b_{2n}+b_{2n+1}=2a_n+1$

    을 이고 이를 이용하면 되네?

     

    $\begin{array}{rl}\sum _{n=1}^{63}{a}_n& =a_1+(a_2+a_3)+(a_4+a_5)+\cdots +(a_{62}+a_{63})\\ \\ & =a_1+(2b_1+1)+(2b_2+1)+\cdots +(2b_{31}+1)\\ \\ & =a_1+31+2\sum _{n=1}^{31}{b}_n\end{array}$$\begin{aligned} \sum\limits_{ n=1}^{63}a_n &=a_1+(a_2+a_3)+(a_4+a_5)+\cdots +(a_{62}+a_{63})\\ \\ &= a_1 +(2b_1+1)+(2b_2+1)+\cdots +(2b_{31}+1) \\ \\ &= a_1 +31+2\sum\limits_{ n=1}^{31}b_n\end{aligned}$

  • $\sum _{n=1}^{63}{a}_n-\sum _{n=1}^{31}{b}_n=155$$\sum\limits_{ n=1}^{63}a_n-\sum\limits_{ n=1}^{31}b_n =155$에 대입하면,

    $\begin{array}{rl}\sum _{n=1}^{63}{a}_n-\sum _{n=1}^{31}{b}_n& =a_1+31+\sum _{n=1}^{31}{b}_n\\ \\ & =155\end{array}$$\begin{aligned} \sum\limits_{ n=1}^{63}a_n-\sum\limits_{ n=1}^{31}b_n &=a_1+31+\sum\limits_{ n=1}^{31}b_n\\ \\ &=155\end{aligned}$

    $\therefore a_1+\sum _{n=1}^{31}{b}_n=124$$\therefore~ a_1+\sum\limits_{ n=1}^{31}b_n=124$

  • 아무래도 $\sum$$\sum$ 이 $a_n$$a_n$과 $b_n$$b_n$을 와리가리하면서 식이 정리되는 가 보다?

    $\sum _{n=1}^{31}{b}_n=b_1+15+2\sum _{n=1}^{15}{a}_n$$\sum\limits_{ n=1}^{31}b_n= b_1 +15+2\sum\limits_{ n=1}^{15}a_n$

     

    $\sum _{n=1}^{15}{a}_n=a_1+7+2\sum _{n=1}^7{b}_n$$\sum\limits_{ n=1}^{15} a_n=a_1 +7+2\sum\limits_{ n=1}^{7}b_n$

     

    $\sum _{n=1}^7{b}_n=b_1+3+\sum _{n=1}^3{a}_n$$\sum\limits_{ n=1}^7 b_n =b_1+3+\sum\limits_{ n=1}^3 a_n$

     

    $\sum _{n=1}^3{a}_n=a_1+2b_1+1$$\sum\limits_{ n=1}^3a_n=a_1 +2b_1+1$

  정리하면,

  $\begin{array}{rl}{a}_1+\sum _{n=1}^{31}{b}_n& =a_1+b_1+15+2\sum _{n=1}^{15}{a}_n\\ \\ & =3a_1+b_1+15+14+4\sum _{n=1}^7{b}_n\\ \\ & =3a_1+b_1+29+4b_1+12+8\sum _{n=1}^3{a}_n\\ \\ & =11a_1+21b_1+41+8\\ \\ & =11a_1+21b_1+49\\ \\ & =124\end{array}$$\begin{aligned} a_1+\sum\limits_{ n=1}^{31}b_n &=a_1+b_1+15+2\sum\limits_{ n=1}^{15} a_n\\ \\ &= 3a_1+b_1+15+14+4\sum\limits_{ n=1}^7 b_n\\ \\ &= 3a_1+b_1+29 +4b_1 +12+8\sum\limits_{ n=1}^3 a_n\\ \\ &= 11a_1+21b_1 +41+8 \\ \\ &= 11a_1+21b_1+49\\ \\ &=124\end{aligned}$

  $\therefore 11a_1+21b_1=75$$\therefore~ 11a_1+21b_1=75$

• $\{\begin{array}{l}11a_1+21b_1=75\\ \\ b_1=2\end{array}$$\begin{cases} 11a_1+21b_1=75 \\ \\ b_1=2\end{cases}$를 이용하면, $a_1=3$$a_1=3$

• 이제 구하자.

  $b_2=7$$b_2=7$

  $a_4=9$$a_4=9$

  $b_8=25$$b_8=25$

  $a_{16}=27$$a_{16}=27$

  $b_{32}=79$$b_{32}=79$

$\therefore b_{32}=79$$\therefore~ b_{32}=79$