2020학년도 11월 나형 21번

21. 수열 $\{a_n\}$$\big\{a_n\big\}$이 모든 자연수 $n$$n$에 대하여 다음 조건을 만족시킨다.

(가) $a_{2n}=a_n-1$$a_{2n} =a_n -1$

(나) $a_{2n+1}=2a_n+1$$a_{2n+1} =2a_n+1$

$a_{20}=1$$a_{20}=1$일 때, $\sum _{n=1}^{63}{a}_n$$\sum\limits_{ n=1}^{63}a_n$의 값을 구하시오.

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i. 생각

• 우선 조건을 가지고 유추하면서 구할 수 있는 항들을 구해보자.

  $a_{20}=1⟶a_{20}=a_{10}-1⟶a_{10}=2$$a_{20}=1\quad\longrightarrow \quad a_{20}=a_{10}-1\quad \longrightarrow \quad a_{10}=2$

  $a_{10}=2⟶a_{10}=a_5-1⟶a_5=3$$a_{10}=2\quad \longrightarrow \quad a_{10}=a_5-1\quad \longrightarrow \quad a_5=3$

  $a_5=2a_2+1=3⟶a_2=1$$a_5=2a_2+1=3\quad\longrightarrow \quad a_2=1$

  $a_2=a_1-1=1⟶a_1=2$$a_2=a_1-1=1\quad \longrightarrow \quad a_1=2$

• 규칙을 찾는 데 집중하면서 항들을 표현해보자.

  $a_1=2$$a_1=2$

  $a_2=a_1-1$$a_2=a_1-1$

  $a_3=2a_1+1$$a_3=2a_1+1$

  $a_4=a_2-1$$a_4=a_2-1$

  $a_5=2a_2+1$$a_5=2a_2+1$

  $a_6=a_3-1$$a_6=a_3-1$

  $a_7=2a_3+1$$a_7=2a_3+1$

  $⋮$$\quad \quad \vdots$

  $a_{62}=a_{31}-1$$a_{62}=a_{31}-1$

  $a_{63}=2a_{31}+1$$a_{63}=2a_{31}+1$

   

  눈이 뚫어지게 뭔가 없는 가를 살펴보면,

  $a_2+a_3=3a_1$$a_2+a_3=3a_1$

  $a_4+a_5=3a_2$$a_4+a_5=3a_2$

  $a_6+a_7=3a_3$$a_6+a_7=3a_3$

  $⋮$$\quad \quad \vdots$

  $a_{62}+a_{63}=3a_{31}$$a_{62}+a_{63}=3a_{31}$

  오!

  $\sum _{n=1}^{63}{a}_n=a_1+3(a_1+a_2+\cdots +a_{31})$$\sum\limits_{ n=1}^{63}a_n =a_1+3(a_1+a_2+\cdots +a_{31})$

  다시 정리하면,

  $\sum _{n=1}^{63}{a}_n=a_1+3\sum _{n=1}^{31}{a}_n$$\sum\limits_{ n=1}^{63}a_n=a_1+3\sum\limits_{ n=1}^{31}a_n$

  오!

• 찾은 규칙을 가지고 계속 반복해보자.

  $\begin{array}{rl}\sum _{n=1}^{63}{a}_n& =a_1+3\sum _{n=1}^{31}{a}_n\\ \\ & =a_1+3(a_1+3\sum _{n=1}^{15}{a}_n)\\ \\ & =4a_1+9\sum _{n=1}^{15}{a}_n\\ \\ & =4a_1+9(a_1+3\sum _{n=1}^7{a}_n)\\ \\ & =13a_1+27\sum _{n=1}^7{a}_n\\ \\ & =13a_1+27(a_1+3\sum _{n=1}^3{a}_n)\\ & =13a_1+27(a_1+3\cdot 4a_1)\\ \\ & =13a_1+27(13a_1)\\ \\ & =13a_1\times 28\\ \\ & =13\cdot 2\cdot 28\\ \\ & =728\end{array}$$\begin{aligned} \sum\limits_{ n=1}^{63} a_n &=a_1 +3\sum\limits_{n=1}^{31}a_n \\ \\ &= a_1+3\Big( a_1+3\sum\limits_{ n=1}^{15}a_n \Big)\\ \\ &= 4a_1+9\sum\limits_{ n=1}^{15}a_n\\ \\ &=4a_1+9\Big(a_1+3\sum\limits_{ n=1}^7a_n\Big)\\ \\ &=13a_1 +27\sum\limits_{ n=1}^{7}a_n\\ \\ &= 13a_1 +27\Big(a_1 +3\sum\limits_{ n=1}^3a_n\Big) \\ \ &=13a_1+27\big(a_1+3\cdot 4 a_1\big)\\ \\ &= 13a_1 +27(13a_1) \\ \\ &=13a_1\times 28\\ \\ &=13\cdot 2 \cdot 28 \\ \\ &=728 \end{aligned}$