문제 은행 · 기초 수학 · 대수 · 수열

수열의 귀납적 정의

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

Sundstrom, Mathematical Reasoning · 4.3절 · 2f번영어 원문

Assume that f1,f2,…,fn,…f_1 ,f_2 , \ldots ,f_n , \ldots are the Fibonacci numbers. Prove each of the following:

For each n∈Nn \in \mathbb{N}, f12+f22+⋯+fn2=fnfn+1f_1^2 + f_2^2 + \cdots + f_n^2 = f_n f_{n + 1}.

정답·풀이 보기 (영어 원문)

Let P(n)P( n ) be, “f12+f22+⋯+fn2=fnfn+1f_1^2 + f_2^2 + \cdots + f_n^2 = f_n f_{n + 1}.” For the basis step, we notice that f12=1f_1^2 = 1 and f1⋅f2=1f_1 \cdot f_2 = 1 and hence, P(1)P(1) is true. For the inductive step, we need to prove that if P(k)P(k) is true, then P(k+1)P(k+1) is true. That is, we need to prove that if f12+f22+⋯+fk2=fkfk+1f_1^2 + f_2^2 + \cdots + f_k^2 = f_k f_{k + 1}, then f12+f22+⋯+fk2+fk+12=fk+1fk+2f_1^2 + f_2^2 + \cdots + f_k^2 + f_{k+1}^2 = f_{k+1} f_{k + 2}. To do this, we can use

(f12+f22+⋯+fk2)+fk+12=fkfk+1+fk+12f12+f22+⋯+fk2+fk+12=fk+1(fk+fk+1)=fk+1fk+2\begin{aligned}\left( f_1^2 + f_2^2 + \cdots + f_k^2 \right) + f_{k+1}^2 & = f_k f_{k + 1} + f_{k+1}^2 \\ f_1^2 + f_2^2 + \cdots + f_k^2 + f_{k+1}^2 & = f_{k+1} \left( f_k + f_{k+1} \right) \\ & = f_{k+1} f_{k+2}\end{aligned}

.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 4.3절 · 1번영어 원문

For the sequence a0,a1,a2,…,an,…a_{0}, a_1 ,a_2 , \ldots , a_n , \ldots , assume that a0=1a_0 = 1 and that for each n∈N∪{0}n \in \mathbb{N} \cup \left\{ 0 \right\}, an+1=(n+1)ana_{n + 1} = \left( {n + 1} \right)a_n. Use mathematical induction to prove that for each n∈N∪{0}n \in \mathbb{N} \cup \left\{ 0 \right\}, an=n!a_n = n!.

다 풀었으면 정답 확인

Let P(n)P(n) be an=n!a_n = n!. Since a0=1a_0 = 1 and 0!=10 ! = 1, we see that P(0)P(0) is true. For the inductive step, we assume that k∈N∪{0}k \in \mathbb{N} \cup \{0\} and that P(k)P(k) is true or that ak=k!a_k = k!.

ak+1=(k+1)ak=(k+1)k!=(k+1)!\begin{aligned}a_{k+1} & = \left( k + 1 \right)a_k \\ & = \left( k + 1 \right) k! \\ & = \left( k + 1 \right)!\end{aligned}

. This proves the inductive step that if P(k)P(k) is true, then P(k+1)P(k+1) is true.

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