亚理学系列

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While it is possible to add arithmetic series one term at a time, it is not feasible or efficient when there are a large number of terms. What is a clever way to add up all the whole numbers between 1 and 100?
::虽然可以一次增加一个术语的算术序列,但当有大量术语时,这个术语既不可行,也不有效。 什么是将所有1至100之间的数字相加的聪明方法?

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Arithmetic Series
::亚理学系列

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An  arithmetic series  is a sum of numbers whose consecutive terms form an arithmetic . The key to adding up a finite arithmetic series is to pair up the first term with the last term, the second term with the second to last term and so on. The sum of each pair will be equal.
::算术序列是一个数字总和,其连续术语构成算术。将一个有限的算术序列相加的关键是将第一个学期与最后一个学期对齐,第二个学期与第二个学期和最后一个学期对齐,等等。每一学期的总和将相等。

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Take  the first 10 numbers. Note that $\begin{align*}1+10=2+9=3+8=4+7=5+6=11\end{align*}$ . There are 5 pairs of 11 which total 55 and so the sum of the first 10 numbers is 55. 
::取前十位数。 注意 1+10=2+9=3+8=4+7=5+6=11。 共有五对11对55, 因此前十位数的总和是55。

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Consider a generic series:
::考虑一个通用系列:

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$\begin{align*}\sum \limits_{i=1}^n a_i=a_1+a_2+a_3+\cdots a_n\end{align*}$
::ni=1ai=a1+a2+a3+a3aan

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When you pair the first and the last terms and note that  $\begin{align*}a_n=a_1+(n-1)k\end{align*}$ the sum is:
::当对齐第一个和最后一个条件时,并注意a=a1+(n-1)k 和是:

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$\begin{align*}a_1+a_n=a_1+a_1+(n-1)k=2a_1+(n-1)k\end{align*}$
::a1+an=a1+a1+(n-1)k=2a1+(n-1)k

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When you pair up the second and the second to last terms you get the same sum:
::当你们将二等和二等至最后一学期配对时,你们得到的金额是一样的:

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$\begin{align*}a_2+a_{n-1}=(a_1+k)+(a_1+(n-2)k)=2a_1+(n-1)k\end{align*}$
::a2+an-1=(a1+k)+(a1+(n-2k)=2a1+(n-1)k

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The next logical question to ask is: how many pairs are there? If there are  $\begin{align*}n\end{align*}$ terms total then there are exactly  $\begin{align*}\frac{n}{2}\end{align*}$ pairs. If  $\begin{align*}n\end{align*}$ happens to be even then every term will have a partner and  $\begin{align*}\frac{n}{2}\end{align*}$ will be a whole number. If  $\begin{align*}n\end{align*}$ happens to be odd then every term but the middle one will have a partner and  $\begin{align*}\frac{n}{2}\end{align*}$ will include a  $\begin{align*}\frac{1}{2}\end{align*}$ pair that represents the middle term with no partner. Here is the general formula for arithmetic series:
::下一个要问的逻辑问题是: 有多少对? 如果有 n 条件总数, 那么就有 n2 对。 如果有 n 恰好是 n, 那么每个条件都会有一个伴侣, n2 将是整个数字。 如果有 n 碰巧是奇数, 那么每个条件但中间条件将有一个伴侣, n2 将包括代表中期的12对, 没有伴侣。 以下是计算序列的一般公式 :

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$\begin{align*}\sum \limits_{i=1}^n a_i=\frac{n}{2}(2a_1+(n-1)k)\end{align*}$ where  $\begin{align*}k\end{align*}$ is the common difference for the terms in the series.
::ni=1ai=n2( 2a1+(n-1)k), K 是序列中术语的常见差 。

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked how to add up the whole numbers between 1 and 100. Gauss was a mathematician who lived hundreds of years ago and there is an anecdote told about him when he was a young boy in school. When misbehaving, his teacher asked him to add up all the numbers between 1 and 100 and he stated 5050 within a few seconds.
::早些时候,有人问您如何将数字加起来,在1到100之间。 高斯是一位数学家,几百年前就曾生活过,当时有一个小男孩在学校时被告知他的情况。 当他行为不端时,他的老师要求他把所有数字加起来,在1到100之间,他在几秒钟内就说了5050的话。

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You should notice that  $\begin{align*}1+100=2+99=\cdots=101\end{align*}$ and that there are exactly 50 pairs that sum to be 101. $\begin{align*}50 \cdot 101=5050\end{align*}$ .
::您应该注意到 1+100=2+99101, 并且有50对正好是101 5010=50。

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Example 2
::例2

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Evaluate the following sum.
::评估以下金额。

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$\begin{align*}\sum \limits_{k=0}^5 5k-2\end{align*}$
::5k=05k-2

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The first term is -2, the last term is 23 and there are 6 terms making 3 pairs. A common mistake is to forget to count the 0 index .
::第一个词是 - 2, 最后一个词是 23, 6 个词组成三对。 一个常见的错误是忘记计算 0 指数 。

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$\begin{align*}\sum \limits_{k=0}^5 5k-2=\frac{6}{2} \cdot (-2+23)=3 \cdot 21=63\end{align*}$
::5k=05k-2=62(-2+23)=321=63

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Example  3
::例3

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Sum the first 15 terms of the following arithmetic sequence.
::以下算术顺序的前15个条件总和。

$\begin{align*}-1,\frac{2}{3},\frac{7}{3},4,\frac{17}{3} \ldots\end{align*}$

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The initial term is -1 and the common difference is $\begin{align*}\frac{5}{3}\end{align*}$ .
::最初的学期是 - 1,共同的差别是53。

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$$\begin{align*}\sum \limits_{i=1}^n a_i&=\frac{n}{2}(2a_1+(n-1)k) \\ &=\frac{15}{2}\left(2(-1)+(15-1)\frac{5}{3}\right) \\ &=\frac{15}{2}\left(-2+14 \cdot \frac{5}{3}\right) \\ &=160 \end{align*}$$
::ni=1ai=n2( 2a1+(n-1)1k)=152(2(-1)+(15-1)53=152(-2+1453)=160

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Example 4
::例4

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Evaluate the following sum.
::评估以下金额。

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$\begin{align*}\sum \limits_{i=0}^{500} 2i-312\end{align*}$
::500i=02i-312

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The initial term is -312 and the common difference is 2.
::最初的学期是312个,共同差为2个。

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$$\begin{align*}\sum \limits_{i=0}^{500} 2i-312&=\frac{501}{2}(2(-312)+(501-1)2) \\ &=94188\end{align*}$$
::500i=02i-312=5012(2(-312)+(501-1)2=9418)

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Example 5
::例5

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Try to evaluate the sum of the following geometric series using the same technique as you would for an arithmetic series.
::尝试使用与算术序列相同的技术来评估以下几何序列的总和。

$\begin{align*}\frac{1}{8}+\frac{1}{2}+2+8+32\end{align*}$

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The real sum is: $\begin{align*}\frac{341}{8}\end{align*}$
::实际金额是: 3418

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When you try to use the technique used for you get:  $\begin{align*}3 \left(\frac{1}{8}+32\right)=\frac{771}{8}\end{align*}$
::尝试使用该技术时,可获得 3( 3( 18+ 32) = 7718

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It is important to know that geometric series have their own method for summing. The method learned in this concept only works for arithmetic series.
::重要的是,要知道几何序列有其自身的总结方法,在这一概念中学习的方法只能用于算术序列。

Summary

播放段落 An arithmetic series is a sum of numbers whose consecutive terms form an arithmetic sequence. ::算术序列是数字的总和,其连续术语构成算术序列。 播放段落 To add up a finite arithmetic series, pair up the first term with the last term, the second term with the second to last term, and so on, as the sum of each pair will be equal. ::为了加起来一个有限的算术序列,第一个学期加最后一个学期,第二个学期加第二个学期加最后一个学期加最后一个学期加最后一个学期,等等,因为每一对学期的总和是相等的。 播放段落 The sum of the first and last terms is $\begin{align*}(a_1 + a_n),\end{align*}$  and the sum of the second and second to last terms is $\begin{align*}(a_2 + a_{n-1}).\end{align*}$   ::第一个和最后一个任期的总和是(a1+an),第二个和第二个至最后一个任期的总和是(a2+an-1)。 播放段落 The number of pairs in an arithmetic series with n terms is $\begin{align*}\frac{n}{2}\end{align*}$  if $\begin{align*}n\end{align*}$  is even, and $\begin{align*}\frac{n-1}{2}\end{align*}$  if $\begin{align*}n\end{align*}$  is odd. ::以 n 值表示的算术序列中的对数为 n2, 如果 n 是偶数, n 则为n- 12, 如果 n 是奇数 。 播放段落 The general formula for an arithmetic series is $\begin{align*}\sum \limits_{i=1}^n a_i=\frac{n}{2}(2a_1+(n-1)k)\end{align*}$  where $\begin{align*}k\end{align*}$  is the common difference. ::算术序列的一般公式为 ni=1ai=n2( 2a1+(n-1)k),K是共同差。

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Review
::回顾

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1. Sum the first 24 terms of the sequence $\begin{align*}1,5,9,13,\ldots\end{align*}$
::1. 1,5,9,13,...

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2. Sum the first 102 terms of the sequence $\begin{align*}7,9,11,13,\ldots\end{align*}$
::2. 序号7,9,11,13,...

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3. Sum the first 85 terms of the sequence $\begin{align*}-3,-1,1,3,\ldots\end{align*}$
::3. 顺序的前85个条件-3,-1,1,1,3,...

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4. Sum the first 97 terms of the sequence $\begin{align*}\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\ldots\end{align*}$
::4. 顺序13,23,1,43...

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5. Sum the first 56 terms of the sequence  $\begin{align*}- \frac{2}{3},\frac{1}{3},\frac{4}{3},\ldots\end{align*}$
::5. 顺序的第56个条件 - 23,13,43,...

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6. Sum the first 91 terms of the sequence $\begin{align*}-8,-4,0,4,\ldots\end{align*}$
::6. 序号8 - 4,0,4,...

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Evaluate the following sums.
::评估以下金额。

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7. $\begin{align*}\sum \limits_{i=0}^{300} 3i+18\end{align*}$
::7. 300-i=03i+18

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8. $\begin{align*}\sum \limits_{i=0}^{215} 5i+1\end{align*}$
::8. 215i=05i+1

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9. $\begin{align*}\sum \limits_{i=0}^{100} i-15\end{align*}$
::9. 100-i=0-15

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10. $\begin{align*}\sum \limits_{i=0}^{85} -13i+1\end{align*}$
::10. 85-i=0-13i+1

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11. $\begin{align*}\sum \limits_{i=0}^{212} -2i+6\end{align*}$
::11. 212-12-i=0-2i+6

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12. $\begin{align*}\sum \limits_{i=0}^{54} 6i-9\end{align*}$
::12. 54i=06i-9

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13. $\begin{align*}\sum \limits_{i=0}^{167} -5i+3\end{align*}$
::13. 167i=0-5i+3

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14. $\begin{align*}\sum \limits_{i=0}^{341} 6i+102\end{align*}$
::14.341i=06i+102

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15. $\begin{align*}\sum \limits_{i=0}^{452} -7i- \frac{5}{2}\end{align*}$
::15. 452i=0-7i-52

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Review (Answers)
::回顾(答复)

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。