课程： 7.6 感应证明 | The Way To Learn

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感应证明

Proving a theory can be a daunting process. N o matter how many times you try something with the same result, how can you be certain that it will always have the same result, no matter what?
::证明一个理论可能是一个艰巨的过程。无论你多少次尝试同样的结果,你怎么能确定它总是具有同样的结果,不管是什么?

For example, if you were to see someone fill a water balloon with ice water and hold it out the window, you would probably either cringe in anticipation of the shouting below, or eagerly watch, depending on the situation. In either case, your response would be based on the fact that you would be certain that a water balloon would pop on someone's head if dropped out the window onto them. Your certainty would be based on your past experience with water balloons and sidewalks, and you'd very likely be correct, but until the balloon actually hits the target, there isn't any way to be absolutely certain it will break.
::例如,如果你看到有人用冰水填满水气球并把它挡在窗外,你可能会因为下面的呼喊而害怕,或者热切地看着,视具体情况而定。不管哪种情况,你的反应都基于这样的事实:如果从窗户上掉下来,水气球会弹到某人的头部。你的确定性将基于你过去对水气球和人行道的经验,你很可能是对的,但是,在气球真正击中目标之前,没有办法绝对肯定气球会破碎。

In math, situations like this occur a lot. Based on repeated experience, you may develop a rule or shortcut to save time or effort when calculating. However, you may be rightly concerned about using such shortcuts on an important exam. After all, how can you be certain that the shortcut works in every situation?
::在数学中,类似这种情况经常发生。根据反复的经验,您可以在计算时制定规则或捷径来节省时间或努力。然而,您或许有理由担心在一次重要考试中使用这种捷径。毕竟,您如何确定快捷键在每一种情况下都有效?

Proofs by Induction
::上岗证明

In this lesson you will learn about mathematical induction , a method of proof that will allow you to prove that a particular statement is true for all positive integers.
::在这个教训中,你会学到数学感应, 这是一种证明方法, 使你能够证明对所有正整数来说, 特定语句是真实的。

First let's make a guess at a formula that will give us the sum of all the positive integers from 1 to n for any integer n . If we look closely at Gauss’s Formula we used in the last lesson, where the young boy was able to quickly add up all of the numbers between 1 and 100, we can see a general form: there were 100 numbers, hence 50 pairs. So if there were n numbers, there would be ( n /2) pairs. The first and last numbers were 1 and 100. They added together to give us 101. This number was the sum of each pair in the overall sum. So in general, we could add together 1 and n to get the sum of each pair. Therefore we might hypothesize that the sum of the first n positive integers is n (( 1 + n)/2). However, we have not proven that this formula works for all positive integers n . Mathematical induction will allow us to do this.
::首先让我们来猜测一下一个公式,这个公式将给我们从1到n的所有正整数的总和,从1到n的整数 n。如果我们仔细看看高斯的公式,我们在最后一个课中使用的公式,这个小男孩能够迅速将数字加在一起,在1到100之间,我们可以看到一个一般的形式:有100个数字,因此有50对。所以如果有n数字,就会有(n/2)对。第一个和最后一个数字是1和100。它们加在一起是为了给我们101。这个数字是每对总和中的每对总和。因此,我们一般可以一起加1和N,以获得每对的总数。因此我们可以假设第一个正整数的总和是n((1+ n/2)。然而,我们没有证明这个公式对所有正整数都有效。数学的介绍会允许我们这样做。

The overall idea of induction is this: Assume that a statement is true for some arbitrary value of n , and show that if the statement is true for n = k , it must also be true for n = k+1 . This process is used because we can’t actually show it is true for every value. For example, you might show that the above equation is true for n = 100, and then n = 101, and then n = 102, but then what about 103? 104? 500? A million?
::上岗总的想法是:假设声明对n的任意值是真实的,并显示如果声明对n=k是真实的,那么对n=k+1也必须是真实的。这个过程之所以被使用,是因为我们不能实际显示它对所有数值都是真实的。例如,你可能显示上述方程式对n=100是真实的,然后n=101,然后n=102,但是大约103?104?500?100万?

Mathematical induction allows us prove that a statement is true in three steps:
::数学上岗培训可以证明声明在三个步骤中是真实的:

Step 1) The base case : prove that the statement is true for the first value of n . In some cases, this might be n = 0. In the case of the integer sum formula above, we would start with n = 1. Often with induction you may want to expand the first step by showing that the statement is true for several values of n .
::步骤1 基本情况:证明该语句对n的第一个值是真实的。在某些情况下,这可能为n=0。在以上整数和公式中,我们从n=1.开始,通常随着感应,你可能想要通过显示该语句对n几个值是真实的来扩展第一步。

Step 2) The inductive hypothesis : assume that the statement is true for the k ^th value of n . In the case of the integer sum formula, we would state the following: the sum of the first k positive integers is k ((1 + k )/2).
::步骤2 引言假设:假定说明对n. kth值正确。对于整数和公式,我们将说明如下:第一个 k 正数整数的和是 k( 1 + k)/2)。

Step 3) The inductive step : use the inductive hypothesis to show that the statement is true for the k + 1 ^th step. In the case of the integer sum formula, we would prove the following: assuming that the sum of the first k positive integers is k ((1 + k )/2) , the sum of the first k +1 positive integers is (( k + 1) (1 + ( k + 1)/2) .
::缩略语步骤 3 : 使用感想假设来表明 k + 第 1 步的语句是真实的。对于整数和公式,我们将证明如下:假设第一个 k 正数整数的总和是 k( 1 + k) /2 , 第一个 k + 1 正数整数的总和是 ( ( k + 1 ) 1 + ( k + 1 + 1) /2) 。

Carrying out this kind of proof requires that you perform each of these steps., For the third step in particular you must rely on your algebra skills.
::执行这种证明要求您执行每一个步骤。,特别是第三步,您必须依靠代数技能。

Examples
::实例

Example 1
::例1

Earlier, you were asked how you could be certain that a theory always results in the correct answer.
::早些时候,有人问过你,你如何可以确定理论总是得出正确的答案。

Situations like this are custom-made for . If you come up with a shortcut on your math, and want to be absolutely certain it works in every situation, run it through the proof explained in this lesson. If it passes all of the tests, you can be sure it will work with any number you throw at it.
::这样的情况是定制的。如果您在数学上想出一条捷径, 并且想要绝对确定它在每个情况下都有效, 请通过这个课中解释的证明。如果它通过所有测试, 您可以确定它会使用任何数字。

Example 2
::例2

Prove: The sum of the first n positive integers is $\begin{aligned} \frac{n (1 + n)}{2} \end{aligned}$ .
::证明:第一个正数正数整数之和为 n(1+n) 2。

Use the three steps of mathematical induction:
::使用数学感应的三个步骤:

Step 1) The base case:
::步骤1 基本情况:

If n = 1, the entire is just 1 and therefore the sum is 1. Also, $\begin{aligned} \frac{n (1 + n)}{2} = \frac{1 (1 + 1)}{2} = \frac{2}{2} = 1 \end{aligned}$ .
::如果 n = 1, 则整个为 1, 因此总和为 1. 1, n(1+n) 2 = 1(1+1) 2 = 22= 1。

This establishes the base case.
::这样就可以确定基数。

Step 2) Assume that the sum of the first $\begin{aligned} k \end{aligned}$ positive integers is $\begin{aligned} \frac{k (1 + k)}{2} \end{aligned}$ .
::第2步假设第一个 k 正数整数的总和是 k(1+k) 2 。

In other words, assume that $\begin{aligned} 1 + 2 + 3 + \dots + k = \frac{k (1 + k)}{2} \end{aligned}$ .
::换句话说,假设 1+2+3k=k(1+k)2。

Step 3) We must show that the sum of the first $\begin{aligned} k + 1 \end{aligned}$ positive integers is $\begin{aligned} \frac{(k + 1) (1 + (k + 1))}{2} \end{aligned}$ .
::第3步,我们必须证明,第一个 k+1 正数整数的总和是 (k+1) 1+(k+1) 2。

In other words, we must show that $\begin{aligned} 1 + 2 + 3 + \dots + k + (k + 1) = \frac{(k + 1) (1 + (k + 1))}{2} \end{aligned}$ .
::换句话说,我们必须显示 1+2+3k+(k+1)=(k+1)1+1(k+1)2。

There are two key ideas to keep in mind as you are carrying out this step: (1) remember to use the assumption and (2) remember how sums work.
::在你采取这一步骤时,有两个关键的想法需要铭记1) 记得使用这一假设,(2) 记得如何计算。

How does the sum of the first k + 1 integers relate to the sum of the first k integers? To get the sum of the first k + 1 integers we must add up all the integers from 1 to k and then add on k + 1, since the sum of the first k + 1 integers is $\begin{aligned} 1 + 2 + 3 + \dots + k + (k + 1) \end{aligned}$ .
::第一个 k + 1 整数的总和与第一个 k 整数的总和有何关联? 要获得第一个 k + 1整数的总和, 我们必须将所有整数从 1 到 k 相加, 然后加到 k + 1 上, 因为第一个 k + 1 整数的总和是 1+2+3 @ k+( k+1) 。

Now we must use our assumption. Remember that we are assuming that $\begin{aligned} 1 + 2 + 3 + \dots + k = \frac{k (1 + k)}{2} \end{aligned}$ .
::现在我们必须使用我们的假设。记住, 我们假设的是 1+2+3+3@ k=k(1+k) 2 。

Substitute $\begin{aligned} \frac{k (1 + k)}{2} \end{aligned}$ in for $\begin{aligned} 1 + 2 + 3 + \dots + k \end{aligned}$ in our expression above for the sum of the first k + 1 integers.
::以上述表达式中的 1+2+3, 取代 k( 1+k) 2, 和第一个 k + 1 整数。

Now we have the sum of the first k + 1 integers is $\begin{aligned} \frac{k (1 + k)}{2} + (k + 1) \end{aligned}$ .
::现在我们有了第一个 k + 1 整数的总和是 k(1+k) 2+ (k+1)。

Remember that we are trying to show that the sum of the first k + 1 integers is $\begin{aligned} \frac{(k + 1) (1 + (k + 1))}{2} \end{aligned}$ . With some algebraic manipulation, we can show that $\begin{aligned} \frac{k (1 + k)}{2} + (k + 1) = \frac{(k + 1) (1 + (k + 1))}{2} \end{aligned}$ .
::记住, 我们正在尝试显示第一个 k+ 1 整数的总和是 (k+1)( 1+( k+1) 2) 。通过某些代数操控, 我们可以显示 k(1+k) 2+( k+1) = (k+1) 1 (1+( k+1) 2 ) 2 。

See below:
::见下文:

$\begin{aligned} \frac{k (1 + k)}{2} + (k + 1) \end{aligned}$
$\begin{aligned} = \frac{k (k + 1)}{2} + \frac{2 (k + 1)}{2} \end{aligned}$	The common denominator is 2 Add the fractions
$\begin{aligned} = \frac{k (k + 1) + 2 (k + 1)}{2} \end{aligned}$	Simplify the numerator
$\begin{aligned} = \frac{k^{2} + k + 2 k + 2}{2} \end{aligned}$
$\begin{aligned} = \frac{k^{2} + 3 k + 2}{2} \end{aligned}$	Factor the numerator
$\begin{aligned} = \frac{(k + 1) (k + 2)}{2} \end{aligned}$	The term $\begin{aligned} (k + 2) \end{aligned}$ is the same as $\begin{aligned} ((k + 1) + 1) \end{aligned}$
$\begin{aligned} = \frac{(k + 1) ((k + 1) + 1)}{2} \end{aligned}$

We have shown that our formula for the sum of the first n integers is true for n = 1. We have also shown that whenever it is true for n = k it is also true for n = k+1 . Since we know it is true for n = 1, it must therefore be true for n = 2. Similarly, since it is true for n = 2, it must therefore be true for n = 3, and it must therefore be true for n = 4,... You should see that we have proven that the sum of the first n positive integers is $\begin{aligned} \frac{n (1 + n)}{2} \end{aligned}$ for all integer values of n . We can similarly prove a formula for the sum of the first n terms in an arithmetic series.
::我们已表明,我们的第一个n整数总和的公式对n=1是真实的。我们还表明,只要n=k是真实的,n=k+1也是真实的。由于我们知道n=1是真实的,因此n=2必须是真实的。同样,n=2也是真实的,n=3也必须是真实的,n=3也必须是真实的,n=4......你应该看到,我们已经证明,第一个n正整数的总和是n(1+n)2,对于n的所有整数值。我们可以同样地证明,计算序列中的第一个n条件总和的公式。

Example 3
::例3

Prove that the sum of the first n terms of an arithmetic series is $\begin{aligned} S_{n} = \frac{n (a_{1} + a_{n})}{2} \end{aligned}$ where a ₁ is the first term in the series and a _n is the last term.
::证明算术序列第一个 n 条件的总和是 Sn=n(a1+an) 2, 其中 a1 是该序列的第一个任期, a 是最后一个任期。

Recall that in an arithmetic sequence or series, there is a common difference, d , between each term, and that the n ^th term is $\begin{aligned} a_{n} = a_{1} + d (n - 1) \end{aligned}$ We need to keep these ideas in mind in order to complete the proof.
::回顾在算术序列或序列中,每个术语之间有共同的差别, d, 并且 nth 术语是 a=a1+d(n-1),我们需要牢记这些想法,以便完成证明。

Step 1) Base case: if $\begin{aligned} n = 1, \end{aligned}$ then $\begin{aligned} S_{1} = a_{1} \end{aligned}$
::步骤1 基本情况:如果 n=1,则S1=a1

Using the hypothesized formula, we have
::使用假设的公式,我们有

	$\begin{aligned} S_{1} = \frac{1 (a_{1} + a_{1})}{2} = \frac{2 a_{1}}{2} = a_{1} \end{aligned}$

Step 2) Assume that $\begin{aligned} S_{k} = \frac{k (a_{1} + a_{k})}{2} \end{aligned}$
::步骤2 假定Sk=k(a1+ak)2

Step 3) Prove that if our formula for $\begin{aligned} S_{k} \end{aligned}$ is true then $\begin{aligned} S_{k + 1} = \frac{(k + 1) (a_{1} + a_{k + 1})}{2} \end{aligned}$ .
::步骤3,证明如果我们的 Sk 公式是真实的,那么Sk+1=(k+1)(a1+ak+1)2。

We can think of the sum of the first k + 1 terms as the sum of the first k terms, plus the k + 1 term. So we have:
::我们可以把第一个 k+1 条件的总和看作是第一个 k 条件的总和,加上 k+ 1 条件。所以我们有:

$\begin{aligned} S_{k + 1} = S_{k} + a_{k + 1} \end{aligned}$	Add the $\begin{aligned} k + 1 \end{aligned}$ term
$\begin{aligned} = \frac{k (a_{1} + a_{k})}{2} + a_{k + 1} \end{aligned}$	Use the formula for $\begin{aligned} S_{k} \end{aligned}$ from step 2
$\begin{aligned} = \frac{k (a_{1} + a_{k})}{2} + \frac{2 a_{k + 1}}{2} \end{aligned}$	The common denominator is 2
$\begin{aligned} = \frac{k (a_{1} + a_{k}) + 2 a_{k + 1}}{2} \end{aligned}$	Add the fractions
$\begin{aligned} = \frac{k (a_{1} + (a_{1} + (k - 1) d)) + 2 (a_{1} + k d)}{2} \end{aligned}$	Use substitution: remember that $\begin{aligned} a_{m} = a_{1} + (m - 1) d \end{aligned}$ for any positive integer $\begin{aligned} m \end{aligned}$ . So $\begin{aligned} a_{k} = a_{1} + (k - 1) d \end{aligned}$ and $\begin{aligned} a_{k + 1} = a_{1} + (k) d \end{aligned}$
$\begin{aligned} = \frac{k (a_{1} + a_{1} + k d - d) + 2 a_{1} + 2 k d}{2} \end{aligned}$
$\begin{aligned} = \frac{k a_{1} + k a_{1} + k^{2} d - k d + 2 a_{1} + 2 k d}{2} \end{aligned}$	Distribute and combine like terms
$\begin{aligned} = \frac{2 k a_{1} + 2 a_{1} + k^{2} d + k d}{2} \end{aligned}$	Factor by grouping
$\begin{aligned} = \frac{2 a_{1} (k + 1) + k d (k + 1)}{2} \end{aligned}$
$\begin{aligned} = \frac{(k + 1) (2 a_{1} + k d)}{2} \end{aligned}$
$\begin{aligned} = \frac{(k + 1) (a_{1} + a_{1} + k d)}{2} \end{aligned}$	Again, $\begin{aligned} a_{k + 1} = a_{1} + (k) d \end{aligned}$
$\begin{aligned} = \frac{(k + 1) (a_{1} + a_{k + 1})}{2} \end{aligned}$

Example 4
::例4

Use induction to prove that $\begin{aligned} 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} \end{aligned}$ .
::使用上传来证明 12+22+32n2=n(n+1)(2n+1)6。

Step 1) Base case: $\begin{aligned} 1^{2} = 1 \end{aligned}$
::基本情况:12=1

$\begin{aligned} \frac{1 (1 + 1) (2 (1) + 1)}{6} = \frac{2 (3)}{6} = 1 \end{aligned}$

Step 2) Inductive hypothesis: $\begin{aligned} 1^{2} + 2^{2} + 3^{2} + . . . + k^{2} = \frac{k (k + 1) (2 k + 1)}{6} \end{aligned}$
::步骤2 感应假设:12+22+32+...+k2=k(k+1)(2k+1)6

Step 3) Inductive step: show that
::步骤3, 步骤3, 引导步骤:显示

$\begin{aligned} 1^{2} + 2^{2} + 3^{2} + . . . + k^{2} + (k + 1)^{2} = \frac{(k + 1) (k + 1 + 1) (2 (k + 1) + 1)}{6} \end{aligned}$
::12+22+32+...+k2+(k+1)2=(k+1)(k+1)(k+1+1)(2(k+1)6

First, note that $\begin{aligned} \frac{(k + 1) (k + 1 + 1) (2 (k + 1) + 1)}{6} = \frac{(k + 1) (k + 2) (2 k + 3)}{6} \end{aligned}$ .
::首先,请注意 (k+1)(k+1)(k+1+1)(2(k+1)+1)6=(k+1)(k+2)(2k+3)6。

Now we have:
::现在我们已经:

	$\begin{aligned} 1^{2} + 2^{2} + 3^{2} + . . . + k^{2} + (k + 1)^{2} \end{aligned}$
	$\begin{aligned} = \frac{k (k + 1) (2 k + 1)}{6} + (k + 1)^{2} \end{aligned}$
	$\begin{aligned} = \frac{k (k + 1) (2 k + 1) + 6 (k + 1)^{2}}{6} = \frac{(k + 1) [k (2 k + 1) + 6 (k + 1)]}{6} \end{aligned}$
	$\begin{aligned} = \frac{(k + 1) [2 k^{2} + k + 6 k + 6]}{6} = \frac{(k + 1) [2 k^{2} + 7 k + 6]}{6} = \frac{(k + 1) (2 k + 3) (k + 2)}{6} \end{aligned}$

Example 5
::例5

Use induction to prove that $\begin{aligned} 1 + 3 + 5 + \dots + (2 n - 1) = n^{2} \end{aligned}$ .
::使用感应来证明 1+3+5( 2n-1) =n2。

Step 1) Base case: $\begin{aligned} 1 = 1^{2} \end{aligned}$
::基本情况:1=12

Step 2) Inductive hypothesis: assume that $\begin{aligned} 1 + 3 + 5 + . . . + (2 k - 1) = k^{2} \end{aligned}$
::步骤2,引导假设:假设1+3+5+...+(2k-1)=k2

Step 3) Show that $\begin{aligned} 1 + 3 + 5 + . . . + (2 k - 1) + (2 k + 1) = (k + 1)^{2} \end{aligned}$
::步骤3 显示 1+3+5+...+( 2k- 1)+( 2k+1)=( k+1)2

We have: $\begin{aligned} 1 + 3 + 5 + . . . + (2 k - 1) + (2 k + 1) = k^{2} + (2 k + 1) \end{aligned}$
::我们有:1+3+5+...+2k-1+2k+1=k2+2(2k+1)

	$\begin{aligned} = k^{2} + 2 k + 1 \end{aligned}$
	$\begin{aligned} = (k + 1) (k + 1) = (k + 1)^{2} \end{aligned}$

Example 6
::例6

Use induction to prove that $\begin{aligned} 1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = \frac{n^{2} (n + 1)^{2}}{4} \end{aligned}$ .
::使用上岗证明 13+23+33\\\\n3=n2(n+1)24。

Step 1) Base case: $\begin{aligned} 1^{3} = 1 \end{aligned}$
::基本情况:13=1

$\begin{aligned} \frac{1^{2} (1 + 1)^{2}}{4} = \frac{2^{2}}{4} = 1 \end{aligned}$

Step 2) Assume that $\begin{aligned} 1^{3} + 2^{3} + 3^{3} + . . . + k^{3} = \frac{k^{2} (k + 1)^{2}}{4} \end{aligned}$
::步骤2),假设13+23+33+...+k3=k2(k+1)24

Step 3) Show that $\begin{aligned} 1^{3} + 2^{3} + 3^{3} + . . . + k^{3} + (k + 1)^{3} = \frac{(k + 1)^{2} ((k + 1) + 1)^{2}}{4} \end{aligned}$
::步骤3 显示 13+23+33+...+k3+(k+1)3=(k+1)2((k+1)+1)24

First, note that $\begin{aligned} \frac{(k + 1)^{2} ((k + 1) + 1)^{2}}{4} = \frac{(k + 1)^{2} (k + 2)^{2}}{4} \end{aligned}$
::首先,注意 (k+1) 2( (k+1) +1) 24= (k+1) 2(k+2) 24

Now we have:
::现在我们已经:

	$\begin{aligned} 1^{3} + 2^{3} + 3^{3} + . . . + k^{3} + (k + 1)^{3} \end{aligned}$
	$\begin{aligned} = \frac{k^{2} (k + 1)^{2}}{4} + (k + 1)^{3} \end{aligned}$
	$\begin{aligned} = \frac{k^{2} (k + 1)^{2} + 4 (k + 1)^{3}}{4} \end{aligned}$
	$\begin{aligned} = \frac{(k + 1)^{2} [k^{2} + 4 (k + 1)]}{4} = \frac{(k + 1)^{2} [k^{2} + 4 k + 4]}{4} = \frac{(k + 1)^{2} (x + 2)^{2}}{4} \end{aligned}$

Review
::回顾

Use induction to prove the following:
::使用上岗培训证明:

$\begin{aligned} - 15 - 25 - 35 + . . . - 10 k - 5 = k (- 5 k - 10) \end{aligned}$
::-15-25-35+...-10k-5=k(-5k-10)

$\begin{aligned} 1 + 9 + 9^{2} + 9^{3} + . . . + 9^{k} = \frac{9^{k + 1} - 1}{9 - 1} \end{aligned}$
::1+9+92+93+...+9k=9k+1-19-1

$\begin{aligned} 4 + 8 + 12 + . . . + 4 k = 2 k (k + 1) \end{aligned}$
::4+8+12+...+4k=2k(k+1)

$\begin{aligned} 6 + 8 + 10 + . . . + 2 k + 4 = k (k + 5) \end{aligned}$
::6+8+10+...+2k+4=k(k+5)

$\begin{aligned} 1 + 2 + 3 + . . . + k = \frac{1}{2} k (k + 1) \end{aligned}$
::1+2+3+...+k=12k(k+1)

$\begin{aligned} 1 + 6 + 6^{2} + 6^{3} + . . . + 6^{k} = \frac{6^{k + 1} - 1}{6 - 1} \end{aligned}$
::1+6+62+63+...+6k=6k+1-16-1

$\begin{aligned} - 1 - 5 - 9 + . . . - 4 k + 3 = k (- 2 k + 1) \end{aligned}$
::-1-5-9+...-4k+3=k(-2k+1)

$\begin{aligned} 1 + 4 + 4^{2} + 4^{3} + . . . + 4^{k} = \frac{4^{k + 1} - 1}{4 - 1} \end{aligned}$
::1+4+42+43+...+4k=4k+1-14-1

$\begin{aligned} 3 + 6 + 9 + . . . + 3 k = \frac{3}{2} k (k + 1) \end{aligned}$
::3+6+9+...+3k=32k(k+1)

$\begin{aligned} - 1 - 3 - 5 + . . . - 2 k + 1 = k (- k) \end{aligned}$
::-1-3-5+...-2k+1=k(-k)

$\begin{aligned} 1 + 3 + 3^{2} + 3^{3} + . . . + 3^{k} = \frac{3^{k + 1} - 1}{3 - 1} \end{aligned}$
::1+3+32+33+...+3k=3k+1-13-1

$\begin{aligned} 1 + 7 + 7^{2} + 7^{3} + . . . + 7^{n} = \frac{7^{n + 1} - 1}{6} \end{aligned}$
::1+7+72+73+...+7n=7n+1-16

$\begin{aligned} 4 + 8 + 12 + . . . + 4 n = 2 n (n + 1) \end{aligned}$
::4+8+12+...+4n=2n(n+1)

$\begin{aligned} 10 + 18 + 26 + . . . 8 n + 2 = n (4 n + 6) \end{aligned}$
::10+18+26+... 8n+2=n(4n+6)

$\begin{aligned} 6 + 12 + 18 + . . . + 6 n = 3 n (n + 1) \end{aligned}$
::6+12+18+...+6n=3n(n+1)

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章节大纲

常规

感应证明

Proofs by Induction ::上岗证明

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

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