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?

#### Watch This

http://www.youtube.com/watch?v=Dj1JZIdIwwo James Sousa: Arithmetic Series

#### Guidance

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. Consider a generic series:

\begin{align*}\sum \limits_{i=1}^n a_i=a_1+a_2+a_3+\cdots a_n\end{align*}

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:

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

When you pair up the second and the second to last terms you get the same sum:

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

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:

@$\begin{align*}\sum \limits_{i=1}^n a_i=\frac{n}{2}(2a_1+(n-1)k)\end{align*}@$ where @$\begin{align*}k\end{align*}@$ is the common difference for the terms in the series.

**
Example A
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Add up the numbers between one and ten (inclusive) in two ways.

**
Solution:
**
One way to add up lists of numbers is to pair them up for easier mental arithmetic.

@$$\begin{align*}1+2+3+4+5+6+7+8+9+10&=3+7+11+15+19 \\ &=10+26+19 \\ &=36+19 \\ &=55\end{align*}@$$

Another way is to 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.

**
Example B
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Evaluate the following sum.

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

**
Solution:
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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.

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

**
Example C
**

Try to evaluate the sum of the following geometric series using the same technique as you would for an arithmetic series.

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

**
Solution:
**

The real sum is: @$\begin{align*}\frac{341}{8}\end{align*}@$

When you try to use the technique used for arithmetic sequences you get: @$\begin{align*}3 \left(\frac{1}{8}+32\right)=\frac{771}{8}\end{align*}@$

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.

**
Concept Problem Revisited
**

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.

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. @$\begin{align*}50 \cdot 101=5050\end{align*}@$ .

#### Vocabulary

An
**
arithmetic series
**
is a sum of numbers whose consecutive terms form an arithmetic sequence.

#### Guided Practice

1. Sum the first 15 terms of the following arithmetic sequence.

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

2. Sum the first 100 terms of the following arithmetic sequence.

@$\begin{align*}-7,-4,-1,2,5,8,\ldots\end{align*}@$

3. Evaluate the following sum.

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

**
Answers:
**

1. The initial term is -1 and the common difference is @$\begin{align*}\frac{5}{3}\end{align*}@$ .

@$$\begin{align*}\sum \limits_{i=1}^n a_i&=\frac{n}{2}(2a_1+(n-1)k) \\ &=\frac{15}{2}\left(2(-1)+(15-1)\frac{5}{3}\right) \\ &=\frac{15}{2}\left(-2+14 \cdot \frac{5}{3}\right) \\ &=160 \end{align*}@$$

2. The initial term is -7 and the common difference is 3.

@$$\begin{align*}\sum \limits_{i=1}^n a_i&=\frac{n}{2}(2a_1+(n-1)k) \\ &=\frac{100}{2}(2(-7)+(100-1)3) \\ &=14150\end{align*}@$$

3. The initial term is -312 and the common difference is 2.

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

#### Practice

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

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

3. Sum the first 85 terms of the sequence @$\begin{align*}-3,-1,1,3,\ldots\end{align*}@$

4. Sum the first 97 terms of the sequence @$\begin{align*}\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\ldots\end{align*}@$

5. Sum the first 56 terms of the sequence @$\begin{align*}- \frac{2}{3},\frac{1}{3},\frac{4}{3},\ldots\end{align*}@$

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

Evaluate the following sums.

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

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

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

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

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

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

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

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

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