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수학적 귀납법

수학적 귀납법으로 자연수에 대한 명제를 증명한다

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

27문제 · 1 / 2쪽

1.Sundstrom, Mathematical Reasoning · 4.1절 · 3a번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn, 2+5+8+⋯+(3n−1)=n(3n+1)22 + 5 + 8 + \cdots + \left( {3n - 1} \right) = \dfrac{{n\left( {3n + 1} \right)}}{2}.

정답·풀이 보기 (영어 원문)

For each n∈Nn \in \mathbb{N}, let P(n)P( n ) be, 2+5+8+⋯+(3n−1)=n(3n+1)22 + 5 + 8 + \cdots + (3n - 1) = \dfrac{{n( {3n + 1} )}}{2}. When we use n=1n = 1, the summation on the left side of the equation is 2, and the right side is 1(3⋅1+1)2=2\dfrac{1(3 \cdot 1 + 1)}{2} = 2. Therefore, P(1)P( 1 ) is true. For the inductive step, let k∈Nk \in \mathbb{N} and assume that P(k)P \left( k \right) is true. Then,

2+5+8+⋯+(3k−1)=k(3k+1)22 + 5 + 8 + \cdots + \left( {3k - 1} \right) = \frac{{k \left( {3k + 1} \right)}}{2}

. We now add 3(k+1)−13 \left( k + 1 \right) - 1 to both sides of this equation. This gives

2+5+8+⋯+(3k−1)+3(k+1)−1=k(3k+1)2+(3(k+1)−1)=k(3k+1)2+(3k+2)\begin{aligned}2 + 5 + 8 + \cdots + (3k - 1) & \\ + 3 \left( {k + 1} \right)-1 & = \frac{{k \left( {3k + 1} \right)}}{2} + \left( 3 \left( k + 1 \right) - 1 \right) \\ & = \frac{{k \left( {3k + 1} \right)}}{2} + \left( {3k + 2} \right)\end{aligned}

If we now combine the terms on the right side of the equation into a single fraction, we obtain

2+5+⋯+(3k−1)+3(k+1)−1=k(3k+1)+6k+42=3k2+7k+42=(k+1)(3k+4)2=(k+1)(3(k+1)+1)2\begin{aligned}2 + 5 + \cdots + \left( {3k - 1} \right) + 3 \left( {k + 1} \right)-1 & = \frac{k \left( 3k + 1 \right) + 6k+4}{2} \\ & = \frac{ 3k^2 + 7k + 4}{2} \\ & = \frac{ \left( k + 1 \right) \left( 3k + 4 \right)}{2} \\ & = \frac{ \left( k + 1 \right) \left( 3 \left( k + 1 \right) + 1 \right)}{2}\end{aligned}

This proves that if P(k)P\left( k \right) is true, then P(k+1)P\left( {k + 1} \right) is true.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 2b번
# 수학적 귀납법증명·논술자동 분류
2.Sundstrom, Mathematical Reasoning · 4.1절 · 3b번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn, 1+5+9+⋯+(4n−3)=n(2n−1)1 + 5 + 9 + \cdots + \left(4n - 3 \right) = n \left(2n - 1 \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
3.Sundstrom, Mathematical Reasoning · 4.1절 · 3c번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn, 13+23+33+⋯+n3=[n(n+1)2]21^3 + 2^3 + 3^3 + \cdots + n^3 = \left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 8번
# 수학적 귀납법증명·논술자동 분류
4.Sundstrom, Mathematical Reasoning · 4.1절 · 8a번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn, 3 divides (4n−1)\left( {4^n - 1} \right).

정답·풀이 보기 (영어 원문)

The key to the inductive step is that if 4k=1+3m4^k = 1 + 3m, then 4k⋅4=4(1+3m)4^k \cdot 4 = 4 ( 1 + 3m ), which implies that

4k+1−1=3(1+4m)4^{k+1} - 1 = 3 (1 + 4m )

.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
5.Sundstrom, Mathematical Reasoning · 4.1절 · 8b번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn, 6 divides (n3−n)\left( {n^3 - n} \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 8번
# 수학적 귀납법증명·논술자동 분류
6.Sundstrom, Mathematical Reasoning · 4.1절 · 14번영어 원문

Use mathematical induction to prove that the sum of the cubes of any three consecutive natural numbers is a multiple of 9.

정답·풀이 보기 (영어 원문)

Three consecutive natural numbers may be represent by nn, n+1n + 1, and n+2n + 2, where nn is a natural number. For the inductive step, think before you try to do a lot of algebra. You should be able to complete a proof of the inductive step by expanding the cube of only one expression.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
7.Sundstrom, Mathematical Reasoning · 4.1절 · 16a번영어 원문

In calculus, it can be shown that

∫sin⁡ 2x dx=x2−12sin⁡xcos⁡x+c and ∫cos⁡ 2x dx=x2+12sin⁡xcos⁡x+c\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}

. Using integration by parts, it is also possible to prove that for each natural number nn,

∫sin⁡ nx dx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡ n−2x dx and ∫cos⁡ nx dx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡ n−2x dx\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}

.

Determine the values of

∫0π/2sin⁡ 2x dx and ∫0π/2sin⁡ 4x dx\int_{0}^{\pi/2} \sin^{\,2} x \, dx \qquad \text{ and } \qquad \int_{0}^{\pi/2} \sin^{\,4} x \, dx

.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 8번
# 수학적 귀납법증명·논술자동 분류
8.Sundstrom, Mathematical Reasoning · 4.1절 · 16b번영어 원문

In calculus, it can be shown that

∫sin⁡ 2x dx=x2−12sin⁡xcos⁡x+c and ∫cos⁡ 2x dx=x2+12sin⁡xcos⁡x+c\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}

. Using integration by parts, it is also possible to prove that for each natural number nn,

∫sin⁡ nx dx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡ n−2x dx and ∫cos⁡ nx dx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡ n−2x dx\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}

.

Use mathematical induction to prove that for each natural number nn,

∫0π/2sin⁡ 2nx dx=1⋅3⋅5⋯(2n−1)2⋅4⋅6⋯(2n)π2 and ∫0π/2sin⁡ 2n+1x dx=2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n+1)\begin{aligned}\int_{0}^{\pi/2} \sin^{\,2n}x \, dx & = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \frac{\pi}{2} \text{ and } \\ \int_{0}^{\pi/2} \sin^{\,2n+1}x \, dx & = \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{1 \cdot 3 \cdot 5 \cdots (2n + 1)}\end{aligned}

. These are known as the Wallis sine formulas.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 5번
# 수학적 귀납법증명·논술자동 분류
9.Sundstrom, Mathematical Reasoning · 4.1절 · 16c번영어 원문

In calculus, it can be shown that

∫sin⁡ 2x dx=x2−12sin⁡xcos⁡x+c and ∫cos⁡ 2x dx=x2+12sin⁡xcos⁡x+c\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}

. Using integration by parts, it is also possible to prove that for each natural number nn,

∫sin⁡ nx dx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡ n−2x dx and ∫cos⁡ nx dx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡ n−2x dx\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}

.

Use mathematical induction to prove that

∫0π/2cos⁡ 2nx dx=1⋅3⋅5⋯(2n−1)2⋅4⋅6⋯(2n)π2 and ∫0π/2cos⁡ 2n+1x dx=2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n+1)\begin{aligned}\int_{0}^{\pi/2} \cos^{\,2n}x \, dx & = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \frac{\pi}{2} \text{ and } \\ \int_{0}^{\pi/2} \cos^{\,2n+1}x \, dx & = \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\end{aligned}

. These are known as the Wallis cosine formulas.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 12번
# 수학적 귀납법증명·논술자동 분류
10.Sundstrom, Mathematical Reasoning · 4.1절 · 17a번영어 원문

Complete the following.

Why is it not possible to use mathematical induction to prove a proposition of the form

(∀x∈Q)(P(x))\left( {\forall x \in \mathbb{Q}} \right)\left( {P( x )} \right)

, where P(x)P( x ) is some predicate?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
11.Sundstrom, Mathematical Reasoning · 4.1절 · 17b번영어 원문

Complete the following.

Why is it not possible to use mathematical induction to prove a proposition of the form

For each real number

x

with

x \geq 1

,

P( x )

, where

P( x )

is some predicate?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
12.Sundstrom, Mathematical Reasoning · 4.2절영어 원문

Consider the following question:

For which natural numbers

n

do there exist nonnegative integers

x

and

y

such that

n = 3x + 5y

?

To help answer this question, we will let Z∗={x∈Z∣x≥0}\mathbb{Z}^* = \left\{ {x \in \mathbb{Z} \mid x \geq 0} \right\}, and let P(n)P \left( n \right) be There exist x,y∈Z∗x,y \in \mathbb{Z}^* such that n=3x+5yn = 3x + 5y. Notice that P(1)P(1) is false since if both xx and yy are zero, then 3x+5y=03x + 5y = 0 and if either x>0x > 0 or y>0y > 0, then 3x+5y≥33x + 5y \geq 3. Also notice that P(6)P(6) is true since 6=3⋅2+5⋅06 = 3 \cdot 2 + 5 \cdot 0 and P(8)P(8) is true since 8=3⋅1+5⋅18 = 3 \cdot 1 + 5 \cdot 1.

We could continue trying to determine other values of nn for which P(n)P(n) is true. However, let us see if we can use the work in part (2) to determine if P(13)P(13) is true. Notice that 13=3+1013 = 3 + 10 and we know that P(10)P(10) is true. We should be able to use this to prove that P(13)P(13) is true. This is formalized in the next part.

Let k∈Nk \in \mathbb{N} with k≥10k \geq 10. Prove that if P(8)P( 8 ), P(9)P ( 9 ), …\ldots, P(k)P( k ) are true, then P(k+1)P( {k+1} ) is true.

Hint: k+1=3+(k−2)k + 1 = 3 + \left( {k - 2} \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 12번
# 수학적 귀납법증명·논술자동 분류
13.Sundstrom, Mathematical Reasoning · 4.2절 · 1b번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn with n≥6n \geq 6, 2n>(n+1)22^n > \left( n + 1 \right)^2.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 8a번
# 수학적 귀납법증명·논술자동 분류
14.Sundstrom, Mathematical Reasoning · 4.2절 · 1c번영어 원문

Use mathematical induction to prove each of the following:

For each natural number nn with n≥3n \geq 3, (1+1n)n<n\left( {1 + \dfrac{1}{n}} \right)^n \lt n.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 2a번
# 수학적 귀납법증명·논술자동 분류
16.Sundstrom, Mathematical Reasoning · 4.2절 · 4a번영어 원문

Complete the following.

Verify that (1−14)=34\left( {1 - \dfrac{1}{4}} \right) = \dfrac{3} {4} and that (1−14)(1−19)=46\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) = \dfrac{4} {6}.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 14번
# 수학적 귀납법증명·논술자동 분류
17.Sundstrom, Mathematical Reasoning · 4.2절 · 4b번영어 원문

Complete the following.

Verify that (1−14)(1−19)(1−116)=58\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) \left( {1 - \dfrac{1}{{16}}} \right) = \dfrac{5}{8} and that (1−14)(1−19)(1−116)(1−125)=610\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) \left( {1 - \dfrac{1}{{16}}} \right)\left( {1 - \dfrac{1}{{25}}} \right) = \dfrac{6}{{10}}.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절
# 수학적 귀납법증명·논술자동 분류
18.Sundstrom, Mathematical Reasoning · 4.2절 · 6b번영어 원문

Let y=ln⁡xy = \ln x.

Let nn be a natural number. Formulate a conjecture for a formula for dnydxn\dfrac{{d^n y}}{{dx^n }}. Then use mathematical induction to prove your conjecture.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 2a번
# 수학적 귀납법증명·논술자동 분류
19.Sundstrom, Mathematical Reasoning · 4.2절 · 7번영어 원문

For which natural numbers nn do there exist nonnegative integers xx and yy such that n=4x+5yn = 4x + 5y? Justify your conclusion.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.1절 · 14번
# 수학적 귀납법증명·논술자동 분류
20.Sundstrom, Mathematical Reasoning · 4.2절 · 8번영어 원문

Can each natural number greater than or equal to 4 be written as the sum of at least two natural numbers, each of which is a 2 or a 3? Justify your conclusion. For example, 7=2+2+37 = 2 + 2 + 3, and 17=2+2+2+2+3+3+317 = 2 + 2 + 2 + 2 + 3 + 3 + 3.

정답·풀이 보기 (영어 원문)

Let P(n)P( n ) be, “The natural number nn can be written as a sum of natural numbers, each of which is a 2 or a 3.” Verify that P(4),P(5),P(6)P( 4 ), P( 5 ), P( 6 ), and P(7)P( 7 ) are true. To use the Second Principle of Mathematical Induction, assume that k∈N,k≥5k \in \mathbb{N}, k \geq 5 and that P(4),P(5),…,P(k)P( 4 ), P( 5 ), \ldots , P( k ) are true. Then notice that

k+1=(k−1)+2k + 1 = ( {k - 1} ) + 2

. Since k−1≥4k - 1 \geq 4, we have assumed that P(k−1)P( {k - 1} ) is true. This means that (k−1)(k-1) can be written as a sum of natural numbers, each of which is a 2 or a 3. Since k+1=(k−1)+2k + 1 = (k - 1) + 2, we can conclude that (k+1)(k+1) can be written as a sum of natural numbers, each of which is a 2 or a 3. This completes the proof of the inductive step.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 4.2절 · 5번
# 수학적 귀납법증명·논술자동 분류