13.7 上岗证明

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  上岗证明

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  Introduction
  ::导言

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  Induction is one of many methods for proving mathematical statements about numbers. The basic idea is that you prove a statement is true for a small number like 1. This is called the base case. You show that if the statement is true for some random number $\begin{array}{r}k\end{array}$ , then it must also be true for $\begin{array}{r}k+1\end{array}$ .
  ::上导是证明数字数学语句的多种方法之一。 基本的想法是您证明一个语句对像 1 这样的小数字来说是真实的。 这被称为基数 。 您可以显示, 如果语句对某个随机数字 k 是真实的, 那么对于 k+ 1 也必须是真实的 。

  

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   An induction proof is like dominoes set up in a line, where the base case starts the falling cascade of truth. Once you have shown that, in general, if the statement is true for $\begin{array}{r}k\end{array}$ , then it must also be true for $\begin{array}{r}k+1\end{array}$ . It means that once you show the statement is true for 1, then it must also be true for 2. Then it must also be true for 3, and it must also be true for 4, and so on.
  ::上岗证明就像在一条线上设置的多米诺斯, 基案开始的就是真理的串联。 一旦你证明, 一般来说, 如果声明对 k 是真实的, 那么k+1 也必须是真实的 。 这意味着, 一旦你展示声明对 1 是真实的, 那么它对于 2 也必须是真实的 。 那么, 基案对 3 也必须是真实的, 它对 4 等也必须是真实的 。

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  What happens when you forget the base case?  
  ::当你忘记基本情况时会怎样?

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  Mathematical  Induction
  ::数学上岗

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  Simple mathematical induction has three steps:
  ::简单的数学上岗有三个步骤:

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  1. The base case is where the statement is shown to be true for $\begin{array}{r}n\end{array}$  equal to some small number like 1. Note that sometimes more than one base case is needed. 
     ::基本情况是,n的语句显示为正确,n等于象1. 这样的小数字,请注意,有时需要不止一个基数。
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  2. The inductive hypothesis is where the statement is assumed to be true for $\begin{array}{r}n=k\end{array}$ .
     ::感想假设是声明被认为对n=k是真实的。
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  3. The inductive step is where you show that the statement must be true for $\begin{array}{r}n=k+1\end{array}$ .
     ::感想步骤是您要显示该语句对 n=k+1 的语句必须是真实的 。

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  These three logical pieces will show that the statement is true for every number greater than the base case. In strong mathematical induction, the inductive hypothesis assumes all the statements from $\begin{array}{r}n=1\end{array}$  to  $\begin{array}{r}k\end{array}$  for some integer $\begin{array}{r}k.\end{array}$
  ::这三个逻辑部分将显示该语句对大于基数的每一个数字都是真实的。 在强烈的数学感应中,感应假设从 n=1 到 k 的语句都假定为某些整数 k 。

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  Suppose you want to use induction to prove for  $\begin{array}{r}n\ge 1,2+2^2+2^3+\dots +2^n=2^{n-1}-2\end{array}$ .
  ::假设您想要使用感应来证明 n1, 2+22+23+...+2n=2n- 1-2 。

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  Start with the base case . Show the statement works when $\begin{array}{r}n=1\end{array}$ .
  

  $$\begin{array}{r}{2}^1=2\text{ and }{2}^{1+1}-2=4-2=2.\end{array}$$

  Therefore, $\begin{array}{r}{2}^1=2^{1+1}-2\end{array}$ . (Both sides are equal to 2.)
  ::以基本大小写开始。 显示语句在 n=1. 1 21=2 和 21+1-2=4-2=2. 因此, 21=21+1-2 =2. (两边等于 2) 时起作用 。

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  Next, state your inductive hypothesis . Assume the statement works for $\begin{array}{r}n\end{array}$  equal to some number $\begin{array}{r}k\end{array}$ : 

  $$\begin{array}{r}2+2^2+\cdots +2^k=2^{k+1}-2.\end{array}$$

  (You are assuming this is a true statement.)
  ::接下来,请说明你的感性假设。假设声明的n 等于某些数字 k: 2+222k=2k+1-2。 (你假设这是一个真实的声明。 )

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  Finally , u se algebra to manipulate the $\begin{array}{r}n=k\end{array}$  statement to prove  the statement is also true for $\begin{array}{r}n=k+1\end{array}$ . So, you will be trying to show that  $\begin{array}{r}2+2^2+\cdots +2^{k+1}=2^{(k+1)+1}-2\end{array}$ . Start with the inductive hypothesis, and multiply both sides of the equation by 2. Then, do some algebra to get the equation looking like you want.
  ::最后,使用代数来操控 n=k 语句以证明语句对 n=k+1 也是真实的。 所以, 您将尝试显示 2+222k+1=2( 2k+1)+1-2。 从感应假设开始, 并将方程的两边乘以 2。 然后, 做一些代数来让方程看起来像您想要的一样 。

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  Inductive Hypothesis (starting equation): $\begin{array}{r}2+2^2+\cdots 2^k=2^{k+1}-2\end{array}$
  ::诱导假设(启动方程):2+222k=2k+1-2

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  Multiply by 2: $\begin{array}{r}2(2+2^2+\cdots +2^k)=2(2^{k+1}-2)\end{array}$
  ::乘以 2 2: 2 2+22 2k) = 2 2k+1-2)

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  Rewrite: $\begin{array}{r}{2}^2+2^3+\cdots +2^{k+1}=2^{k+1+1}-4\end{array}$
  ::重写: 22+232k+1=2k+1+1- 4

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  Add 2 to both sides: $\begin{array}{r}2+2^2+2^3+\cdots +2^{k+1}=2^{k+1+1}-4+2\end{array}$
  ::双方加2:2+22+23+23+2k+1=2k+1+1+1+4+2

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  Simplify: $\begin{array}{r}2+2^2+\cdots +2^{k+1}=2^{k+1+1}-2\end{array}$
  ::简化: 2+222k+1=2k+1+1+1-2

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  This is exactly what you were trying to prove! So, first you showed that the statement worked for $\begin{array}{r}n=1\end{array}$ . Then you showed that if the statement works for one number, it must work for the next number. This means the statement must be true for all natural numbers greater than or equal to 1.
  ::这正是你试图证明的! 所以, 首先你展示了该语句为 n=1 工作 。 然后您展示了如果该语句为一个数字工作, 它必须为下一个数字工作。 这意味着该语句必须对所有大于或等于 1 的自然数字都适用。 这意味着该语句必须是真实的。

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  The idea of induction can be hard to understand at first, and it definitely takes practice. One thing that makes induction tricky is that there is not a clear procedure for the "proof" part. With practice, you will start to see some common algebra techniques for manipulating equations to prove what you are trying to prove.
  ::初始化的想法一开始可能很难理解,而且肯定需要实践。 诱导问题很难解决的一件事是,“ 校对” 部分没有明确的程序。 通过实践,你将开始看到一些通用的代数技术来操纵方程式来证明你试图证明什么。

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  The following video explains how to prove a mathematical statement using , and shows two specific  examples:
  ::以下影片解释如何使用数学语句证明, 并展示两个具体例子:

   

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  Examples
  ::实例

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  Example 1
  ::例1

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  There is something wrong with this proof. Can you explain what the mistake is? 
  ::这个证据有问题 你能解释一下错误是什么吗?

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  $\begin{array}{r}Forn\ge 1:1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\end{array}$
  ::n1: 12+22+32n2=n(n+1)(2n+1)6

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  Base Case:  When  $\begin{array}{r}n=1\end{array}$ ,  $\begin{array}{r}1=1^2=\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot 2\cdot 3}{6}=\frac{6}{6}=1\end{array}$ .
  ::基数: 当 n=1, 1=12=1(1+1)(21+1), 6=1236=66=1时 。

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  Inductive Hypothesis: Assume the following statement is true for $\begin{array}{r}n=k\end{array}$ :
  ::推论假说:n=k的假设如下:

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  $$\begin{array}{r}{1}^2+2^2+3^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}.\end{array}$$

  
  ::12+22+32k2=k(k+1)(2k+1)6.

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  Proof: You want to show the statement is true for $\begin{array}{r}n=k+1\end{array}$ .
  ::证据: 您想要显示 n=k+1 的语句是真实的 。

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  “Since the statement is assumed true for $\begin{array}{r}n=k\end{array}$ , which is any number, then it must be true for $\begin{array}{r}n=k+1\end{array}$ . You can just substitute $\begin{array}{r}k+1\end{array}$ in.”
  ::“既然声明对n=k(是任意数字)是真实的,那么对于n=k+1(n=k+1)则必须是真实的。”

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  $\begin{array}{r}{1}^2+2^2+3^2+\cdots +(k+1{)}^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\end{array}$
  ::12+22+32(k+1)2=(k+1)((k+1)+1)(2(k+1)+1)(2(k+1)+1)6

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  Solution:  
  ::解决方案 :

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  This is the most common fallacy when doing induction proofs. The fact that the statement is assumed to be true for $\begin{array}{r}n=k\end{array}$ does not immediately imply that it is true for $\begin{array}{r}n=k+1,\end{array}$ and you cannot just substitute in $\begin{array}{r}k+1\end{array}$ to produce what you are trying to show. This is equivalent to assuming true for all numbers and then concluding true for all numbers, which is circular and illogical.
  ::这是做上岗证明时最常见的谬误。 假设n=k的语句是真实的这一事实并不立即意味着n=k+1的语句是真实的, 而且您不能仅仅用 k+1 来替代您想要显示的内容。 这相当于假设所有数字都是真实的, 然后对所有数字来说都是真实的, 这是循环的, 不合逻辑的 。

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  Example 2
  ::例2

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  Write the base case, inductive hypothesis, and what you are trying to show for the statement below. Do not actually prove it.
  ::写基本案例, 感想假设, 以及您试图为下面的语句显示的内容。 不要实际证明它 。

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  $$\begin{array}{r}{1}^3+2^3+3^3+\cdots +n^3=\frac{n^2(n+1{)}^2}{4}\end{array}$$

  
  ::13+23+33n3=n2(n+1)24

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  Solution:
  ::解决方案 :

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  Base Case:  When  $\begin{array}{r}n=1\end{array}$ ,  $\begin{array}{r}{1}^3=\frac{1^2(1+1{)}^2}{4}.\end{array}$  (Both sides are equal to 1.)
  ::基本情况:n=1, 13=12(1+1)24。 (两边等于1)

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  Inductive Hypothesis:  Assume the following statement is true for $\begin{array}{r}n=k\end{array}$ :
  ::推论假说:n=k的假设如下:

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  $$\begin{array}{r}{1}^3+2^3+3^3+\cdots +k^3=\frac{k^2(k+1{)}^2}{4}.\end{array}$$

  
  ::13+23+33k3=k2(k+1)24。

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  Proof:  Prove that the following is true for $\begin{array}{r}n=k+1\end{array}$ :
  ::证据:证明n=k+1的下列情况属实:

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  $$\begin{array}{r}{1}^3+2^3+3^3+\cdots +k^3+(k+1{)}^3=\frac{(k+1{)}^2((k+1)+1{)}^2}{4}.\end{array}$$

  
  ::13+23+33k3+(k+1)3=(k+1)((k+1)2((k+1)+1)24。

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  Example 3
  ::例3

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  Prove the following statement: $\begin{array}{r}\text{For }n\ge 1,1^3+2^3+3^3+\dots +n^3=(1+2+3+\dots +n{)}^2.\end{array}$
  ::证明以下语句: n1,13+23+33+...+n3=(1+2+3+...+n)2。

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  Solution:
  ::解决方案 :

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  Base Case(s):  Two base cases are shown; however, only one is actually necessary.  
  ::基本案例:显示两个基本案例;然而,实际上只需要一个案例。

  $$\begin{array}{rl}{1}^3& =1^2\\ 1^3+2^3& =1+8=9=3^2=(1+2{)}^2\end{array}$$

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  Inductive Hypothesis: Assume the statement is true for some number  $\begin{array}{r}n=k\end{array}$ . In other words, assume the following is true:
  ::推论假说:假设一些n=k的语句是真实的。换句话说,假设以下是真实的:

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  $$\begin{array}{r}{1}^3+2^3+3^3+\dots +k^3=(1+2+3+\dots +k{)}^2.\end{array}$$

  
  ::13+23+33+...+k3=(1+2+3+...+k)2.

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  Proof: You want to show the statement is true for $\begin{array}{r}n=k+1\end{array}$ . It is a good idea to restate what your goal is at this point. Your goal is to show that
  ::证明: 您想要显示对 n=k+1 的语句是真实的 。 最好在此重述您的目标。 您的目标是要显示

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  $$\begin{array}{r}{1}^3+2^3+3^3+\dots +k^3+(k+1{)}^3=(1+2+3+\dots +k+(k+1){)}^2.\end{array}$$

  
  ::13+23+33+...+k3+(k+1)3=(1+2+3+...+k+(k+1))2。

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  You need to start with the assumed case and do algebraic manipulations until you have created what you are trying to show (the equation above):
  ::您需要从假设的大小写开始, 并进行代数操纵, 直到您创建了您想要显示的东西( 以上方程式) :

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  $$\begin{array}{r}{1}^3+2^3+3^3+\dots +k^3=(1+2+3+\dots +k{)}^2.\end{array}$$

  
  ::13+23+33+...+k3=(1+2+3+...+k)2.

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  From the work you have done with arithmetic series, you should notice
  ::从你对计算序列所完成的工作来看,你应该注意到

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  $$\begin{array}{r}1+2+3+4+\cdots +k=\frac{k}{2}(2+(k-1))=\frac{k(k+1)}{2}.\end{array}$$

  
  ::1+2+3+4k=k2(2+(k-1))=k(k+1)2。

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  Substitute into the right side of the equation, and add $\begin{array}{r}(k+1{)}^3\end{array}$ to both sides:
  ::替代方程式的右侧, 并在两侧添加( k+1) 3:

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  $$\begin{array}{r}{1}^3+2^3+3^3+\dots +k^3+(k+1{)}^3={\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3.\end{array}$$

  
  ::13+23+33+...+k3+(k+1)3=(k(k+1)2+(k+1)3。

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  When you combine the righthand side algebraically, you get the result of another arithmetic series.
  ::当将右手边的代数组合起来时,就会得到另一个算术序列的结果。

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  $$\begin{array}{r}{1}^3+2^3+3^3+\dots +k^3+(k+1{)}^3={\left(\frac{(k+1)(k+2)}{2}\right)}^2=(1+2+3+\cdots k+(k+1){)}^2\end{array}$$

  
  ::13+23+33+...+k3+(k+1)3=((k+1)(k+2)(k+2)2)2=(1+2+2+3@k+1)2=(1+2+3@k+1)2

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  Example 4
  ::例4

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  Recall the problem from the Introduction: What happens when you forget the base case? 
  ::回顾导言中的问题:忘记基本案例会怎样?

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  Solution:
  ::解决方案 :

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  If you forget the base case in an induction proof, then you haven't really proved anything. You can get silly results like this "proof" of the statement: " $\begin{array}{r}1=3\end{array}$ ."
  ::如果你忘记了上岗证明中的基本案例, 那么您还没有真正证明什么。 您可以得到一些愚蠢的结果, 比如“ 1=3 ” 的“ 校对 ” 。

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  Base Case: Missing.
  ::基本案件:失踪。

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  Inductive Hypothesis: $\begin{array}{r}k=k+1,\end{array}$ where $\begin{array}{r}k\end{array}$ is a counting number.
  ::感想假设: k=k+1, K是数字 。

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  Proof: Start with the assumption step and add 1 to both sides.
  ::证据:从假设步骤开始,向双方增加一个。

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  $$\begin{array}{rl}k& =k+1\\ k+1& =k+2\end{array}$$

  
  ::k=k+1k+1=k+2

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  Thus, by transitivity of equality:
  ::因此,通过平等的过渡性:

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  $$\begin{array}{rl}k& =k+1=k+2\\ k& =k+2\end{array}$$

  
  ::k=k+1=k+2k=k+2

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  Since $\begin{array}{r}k\end{array}$  is a counting number,  $\begin{array}{r}k\end{array}$ could equal 1. Therefore,  $\begin{array}{r}1=3\end{array}$ .
  ::K是数字,K等于1,因此,1=3。

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  Example 5
  ::例5

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  Write the base case, inductive hypothesis, and what you are trying to show for the statement below. Do not actually prove it.  $\begin{array}{r}{n}^3+2n\end{array}$  is divisible by 3 for any positive integer $\begin{array}{r}n\end{array}$ . 
  ::写基本大小写, 缩略假设, 以及您试图为下面的语句显示的内容 。 不要实际证明它。 n3+2n 对任何正整数 n 来说, 3 可除以 3 。

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  Solution:
  ::解决方案 :

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  Base Case: When  $\begin{array}{r}n=1\end{array}$ ,    $\begin{array}{r}{1}^3+2\cdot 1=3,\end{array}$  which is divisible by 3.
  ::基数:n=1, 13+21=3, 可除以3。

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  Inductive Step: Assume the following is true for $\begin{array}{r}n=k\end{array}$ :  $\begin{array}{r}{k}^3+2k\end{array}$ divisible by 3.
  ::感应步骤: 假设n=k: k3+2k 可除以 3

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  Proof: Show the following is true for $\begin{array}{r}n=k+1\end{array}$ :  $\begin{array}{r}(k+1{)}^3+2(k+1)\end{array}$ is divisible by 3.
  ::证明: 显示 n=k+1 的下列情况为真实值k+1) 3+2(k+1) 可除以 3 。

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  Example 6
  ::例6

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  Complete the proof for the previous problem.
  ::完成上一个问题的证据 。

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  Solution:  
  ::解决方案 :

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  The goal is to show that $\begin{array}{r}(k+1{)}^3+2(k+1)\end{array}$ is divisible by 3 if you already know $\begin{array}{r}{k}^3+2k\end{array}$ is divisible by 3. Expand $\begin{array}{r}(k+1{)}^3+2(k+1)\end{array}$ to see what you get:
  ::目标是显示 (k+1) 3+2(k)+2(k+1) 可除以 3, 如果您已经知道 k3+2k 可除以 3. 扩展 (k+1) 3+2(k+1) 以查看您得到什么 :

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  $$\begin{array}{rl}(k+1{)}^3+2(k+1)& =k^3+3k^2+3k+1+2k+2\\ & =(k^3+2k)+3(k^2+k+1)\end{array}$$

  
  :k+1)3+2(k+1)=k3+3k2+3k2+3k+1+2k+2=(k3+2k)+3(k2+K+1)

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  $\begin{array}{r}{k}^3+2k\end{array}$ is divisible by 3 by assumption (the inductive step), and $\begin{array}{r}3(k^2+k+1)\end{array}$ is clearly a multiple of 3, so it is divisible by 3. The sum of two numbers that are divisible by 3 is also divisible by 3.
  ::k3+2k 通过假设(感应步骤)可以3除以3, 3(k2+k+1)明显是3的倍数, 因此可以3除以3。 两个数字加3可以3除以2, 两者之和也可以3除以3。

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  Example 7
  ::例7

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  Prove the following statement using induction:
  ::使用上岗培训证明以下陈述:

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  For $\begin{array}{r}n\ge 1,1+2+3+4+\cdots +n=\frac{n(n+1)}{2}\end{array}$ .
  ::n1, 1+2+3+4n=n( n+1) 2。

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  Solution:
  ::解决方案 :

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  Base Case:  When  $\begin{array}{r}n=1\end{array}$ ,  $\begin{array}{r}1=\frac{1(1+1)}{2}=1\cdot \frac{2}{2}=1\end{array}$  .
  ::基数: 当 n=1, 1=1(1+1) 2=122=1时 。

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  Inductive Hypothesis:   Assume the following is true for $\begin{array}{r}n=k\end{array}$ : 

  $$\begin{array}{r}1+2+3+4+\cdots +k=\frac{k(k+1)}{2}.\end{array}$$

  Proof: Start with what you know, and work to show it true for $\begin{array}{r}n=k+1\end{array}$ .
  ::感知假说: 假设n=k: 1+2+3+4+4@k=k( k+1) 符合以下假设。 Proof: 以您所知道的开头, 并努力显示 n=k+1 的真实性 。

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  Inductive Hypothesis: $\begin{array}{r}1+2+3+4+\cdots +k=\frac{k(k+1)}{2}\end{array}$ .
  ::感应假设:1+2+3+4k=k(k+1)2。

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  Add $\begin{array}{r}k+1\end{array}$ to both sides: $\begin{array}{r}1+2+3+4+\cdots +k+(k+1)=\frac{k(k+1)}{2}+(k+1)\end{array}$ .
  ::向两边添加 k+1 : 1+2+3+4 @k+( k+1)=k( k+1)2+( k+1) +( k+1)。

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  Find a common denominator for the right side:

  $$\begin{array}{r}1+2+3+4+\cdots +k+(k+1)=\frac{k^2+k}{2}+\frac{2k+2}{2}.\end{array}$$

  
  ::为右侧寻找一个共同分母:1+2+3+4k+( k+1)=k2+k2+2k+2k+22。

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  Simplify the right side: $\begin{array}{r}1+2+3+4+\cdots +k+(k+1)=\frac{k^2+3k+2}{2}\end{array}$ .
  ::简化右侧:1+2+3+4@k+(k+1)=k2+3k+22。

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  Factor the numerator of the right side: $\begin{array}{r}1+2+3+4+\cdots +k+(k+1)=\frac{(k+1)(k+2)}{2}\end{array}$ .
  ::乘以右侧的分子数:1+2+3+4k+(k+1)=(k+1)(k+2)2。

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  Rewrite the right side: $\begin{array}{r}1+2+3+4+\cdots +k+(k+1)=\frac{(k+1)((k+1)+1)}{2}\end{array}$ .
  ::重写右侧:1+2+3+4k+(k+1)=(k+1)((k+1)+1)2。

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  Summary
  ::摘要

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  • The base case is the anchor step. It is the 1st domino to fall, creating a cascade, and thus proving the statement true for every number greater than the base case.
    ::基数是锚脚步。 这是首个倒塌的多米诺, 创造了一个级联, 从而证明了语句对每个比基数大的数字来说都是真实的 。
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  • The inductive hypothesis is the step where you assume the statement is true for $\begin{array}{r}k\end{array}$ .
    ::推论假设是假设K的说法对K是真实的一步。
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  • The inductive step is the proof . It is when you show the statement is true for  $\begin{array}{r}k+1,\end{array}$ using only the inductive hypothesis and algebra.
    ::感应步骤就是证明。它是当您显示 k+1 的语句是真实的时,只使用感应假设和代数。

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  Review
  ::回顾

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  For each of the following statements: a) show the base case is true; b) state the inductive hypothesis; and c) state what you are trying to prove in the inductive step/proof. Do not prove yet.
  ::对于下列每一陈述a) 显示基本案例是真实的;(b) 说明感应假设;和(c) 说明你试图在感应步骤/证据中证明什么。请不要证明。

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  1. For $\begin{array}{r}n\ge 5,4n<2^n\end{array}$ .
  ::1. n5, 4n < 2n.

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  2. For  $\begin{array}{r}n\ge 1,8^n-3^n\end{array}$  is divisible by 5.
  ::2. n1, 8n-3n 可除以 5 。

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  3. For $\begin{array}{r}n\ge 1,7^n-1\end{array}$ is divisible by 6.
  ::3. n1, 7n-1可除以6。

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  4. For $\begin{array}{r}n\ge 2,n^2\ge 2n\end{array}$ .
  ::4. 注2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2

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  5. For $\begin{array}{r}n\ge 1,4^n+5\end{array}$ is divisible by 3.
  ::5. n1, 4n+5可除以3。

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  6. For $\begin{array}{r}n\ge 1,0^2+1^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\end{array}$ .
  ::6. n1,02+12n2=n(n+1)(2n+1)6。

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  Prove each of the statements below. Use your answers to problems 1-6 to help you get started.
  ::证明下面的每个语句。 使用您对问题1 - 6的回答来帮助您启动 。

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  7. For $\begin{array}{r}n\ge 5,4n<2^n\end{array}$ .
  ::7. n5, 4n < 2n.

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  8. For $\begin{array}{r}n\ge 1,8^n-3^n\end{array}$  is divisible by 5.
  ::8. n1、8n-3n可除以5。

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  9. For $\begin{array}{r}n\ge 1,7^n-1\end{array}$  is divisible by 6.
  ::9. n1, 7n-1可除以6。

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  10. For $\begin{array}{r}n\ge 2,n^2\ge 2n\end{array}$ .
  ::10. n% 2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2,n2

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  11. For $\begin{array}{r}n\ge 1,4^n+5\end{array}$ is divisible by 3.
  ::11. n1, 4n+5可除以3。

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  12. For $\begin{array}{r}n\ge 1,0^2+1^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\end{array}$ .
  ::12. n1,02+12n2=n(n+1)(2n+1)6。

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  13. You should believe the statement below is clearly false. What happens when you try to prove it true by induction?
  ::13. 你应该相信下面的说法显然是虚假的。 当你试图通过诱导来证明它的真实性时会发生什么情况?

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  For $\begin{array}{r}n\ge 2,n^2<n\end{array}$ .
  ::n2, n2 <n.

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  14. Explain why the base case is necessary for proving by induction.
  ::14. 解释为什么通过上岗培训证明基本情况是必要的。

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  15. The principles of inductive proof can be used for other proofs besides proofs about numbers. Can you prove the following statement from geometry using induction?
  ::15. 除了关于数字的证明之外,还可用于其他证据,可以使用感应证据的原则,你能证明以下几何学使用感应的描述吗?

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  The sum of the interior angles of any $\begin{array}{r}n-gon\end{array}$ is  $\begin{array}{r}180(n-2)\end{array}$ for $\begin{array}{r}n\ge 3\end{array}$ .
  ::任何 n-gon 的内角总和为 n- 3 的180(n-2) 。

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  Review (Answers )
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。