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# Sums of Arithmetic Series

## Sum of numbers whose consecutive terms form an arithmetic sequence.

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Arithmetic Series

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?

### Arithmetic Series

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

Take the first 10 numbers. Note that 1+10=2+9=3+8=4+7=5+6=11\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.

Consider a generic series:

i=1nai=a1+a2+a3+an\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 an=a1+(n1)k\begin{align*}a_n=a_1+(n-1)k\end{align*} the sum is:

a1+an=a1+a1+(n1)k=2a1+(n1)k\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:

a2+an1=(a1+k)+(a1+(n2)k)=2a1+(n1)k\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 n\begin{align*}n\end{align*} terms total then there are exactly n2\begin{align*}\frac{n}{2}\end{align*} pairs. If n\begin{align*}n\end{align*} happens to be even then every term will have a partner and n2\begin{align*}\frac{n}{2}\end{align*} will be a whole number. If n\begin{align*}n\end{align*} happens to be odd then every term but the middle one will have a partner and n2\begin{align*}\frac{n}{2}\end{align*} will include a 12\begin{align*}\frac{1}{2}\end{align*} pair that represents the middle term with no partner. Here is the general formula for arithmetic series:

i=1nai=n2(2a1+(n1)k)\begin{align*}\sum \limits_{i=1}^n a_i=\frac{n}{2}(2a_1+(n-1)k)\end{align*} where k\begin{align*}k\end{align*} is the common difference for the terms in the series.

### Examples

#### Example 1

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.

You should notice that 1+100=2+99==101\begin{align*}1+100=2+99=\cdots=101\end{align*} and that there are exactly 50 pairs that sum to be 101. 50101=5050\begin{align*}50 \cdot 101=5050\end{align*}.

#### Example 2

Evaluate the following sum.

k=055k2\begin{align*}\sum \limits_{k=0}^5 5k-2\end{align*}

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.

k=055k2=62(2+23)=321=63\begin{align*}\sum \limits_{k=0}^5 5k-2=\frac{6}{2} \cdot (-2+23)=3 \cdot 21=63\end{align*}

#### Example 3

Sum the first 15 terms of the following arithmetic sequence.

1,23,73,4,173\begin{align*}-1,\frac{2}{3},\frac{7}{3},4,\frac{17}{3} \ldots\end{align*}

The initial term is -1 and the common difference is 53\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*}

#### Example 4

Evaluate the following sum.

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

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*}

#### Example 5

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*}

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.

### Review

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*}

To see the Review answers, open this PDF file and look for section 12.4.

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### Vocabulary Language: English

arithmetic series

An arithmetic series is the sum of an arithmetic sequence, a sequence with a common difference between each two consecutive terms.

common difference

Every arithmetic sequence has a common or constant difference between consecutive terms. For example: In the sequence 5, 8, 11, 14..., the common difference is "3".

series

A series is the sum of the terms of a sequence.