# 7.2.2: Series Sums and Gauss' Formula

- Page ID
- 14787

## Series Sums and Gauss's Formula

EyeScreem Ltd is having a special sale on their ice cream cones. In order to get as many people in the store to try out their new flavors, they have decided to run the following promotion:

The first customer to buy an ice cream will pay $6 for a large cone with chocolate fudge.

The second customer will pay only $5.90.

The third customer will only pay $5.80.

Each successive ice cream lover will be charged $0.10 less than the last, until everyone that comes in gets a free cone!

How much money will the store bring in during the sale, assuming at least some customers are given free cones?

## Series Sums and Gauss's Formula

### Gauss's Formula

German mathematician C.F. Gauss is often credited with discovering a formula for calculating the sum of a series when he was a young child. The story is likely apocryphal (a legend), but it has been passed down since Gauss lived in the 1700s. According to the story, Gauss’s teacher wanted to occupy students by having them add up large sets of numbers. When Gauss was asked to add up the first 100 integers, he found the sum very quickly, by pairing the numbers:

All of the numbers in the sum could be paired to make groups of 101. There are one hundred numbers being added, so there are such fifty pairs. Therefore the sum is 50(101) = 5050.

The method Gauss used to solve this problem is the basis for a formula that allows us to add together the first * n* positive integers:

\(\ \sum=\frac{(n)(n+1)}{2}\)

### Using a Graphing Calculator

To generate either a sequence or a series, you can use a graphing calculator. The TI-83/84 series gives you several options. You can, for example, work in sequence mode, which allows you to define a sequence and find terms. If you have an explicit formula for a sequence, you can keep your calculator in function mode. For example, consider the sum \(\ \sum_{n=1}^{9} n^{2}\). It would be time consuming to write out the first 9 squares. The calculator is faster. To generate the 9 terms, press <TI font_2nd> [LIST], then select **OPS**, then option 5, **seq**(. This takes you back to the main screen. You should see **seq**(. After this, enter * x*^2,

*, 1, 9, 1). (The*

*x**tells the calculator that*

*x**is the input. The 1 and the 9 tell it the*

*x***limits**of the sum. The second 1 tells the calculator to go up in increments of 1.)

Press <TI font_ENTER>, and you should see the list of squares. Scroll to the right to see all of them. The scrolling will end when you reach 81.

If you want to find the sum of the terms, first store the sequence in a list (see screen below), then <TI font_2nd> [LIST], then select **MATH**, then option 5, **sum(.*** Then* enter the name of the list and press <TI font_)><TI font_ENTER>. You should get 285.

## Examples

Earlier, you were asked to find how much money the store would bring in during its ice cream promotion.

**Solution**

The series sum formula \(\ \sum=\frac{(n)(n+1)}{2}\) is designed for integers, so let's use it to solve for the number of * dimes* brought in (since that is the unit each term reduces by) and then convert to dollars:

\(\ \begin{array}{l}

\sum=\frac{(60)(61)}{2} \\

\sum=\frac{3,660}{2} \\

\sum=1830.00 \rightarrow \$ 183.00

\end{array}\)

The store will bring in $183.00, which will probably * not* cover the costs of the day. However, they will certainly get a lot of people through the door to try out the ice cream!

Expand the **sigma** and find the sum: \(\ \sum_{n=1}^{7}(2 n-3)\).

**Solution**

\(\ \sum_{n=1}^{7}(2 n-3)\) | =(2×1−3)+(2×2−3)+(2×3−3)+(2×4−3)+(2×5−3)+(2×6−3)+(2×7−3) |
---|---|

= (−1)+(1)+(3)+(5)+(7)+(9)+(11) | |

= 35 |

Expand the sigma and find the sum by adding the terms: \(\ \sum_{n=3}^{6}\left(n^{2}-5\right)\).

**Solution**

Why is Gauss's formula not recommended for this question?

\(\ \sum_{n=3}^{6}\left(n^{2}-5\right)\) | \(\ =\left(3^{2}-5\right)+\left(4^{2}-5\right)+\left(5^{2}-5\right)+\left(6^{2}-5\right)\) |
---|---|

= (4)+(11)+(20)+(31) | |

= 66 |

There are actually a couple of reasons not to use Gauss's formula here, but the biggest is that the formula assumes you are adding all of the integers from 0 to the last number in the series. In the question, you are asked to only sum from the 3rd to 6th term.

If the sum of the first * n* integers is 210, what is

*?*

*n***Solution**

Using Gauss's formula:

\(\ \begin{array}{l}

\sum=\frac{(n)(n+1)}{2} \\

210=\frac{(n)(n+1)}{2}

\end{array}\)

Multiply both sides by 2: \(\ 420=(n)(n+1)\)

Distribute: \(\ 420=\left(n^{2}+n\right)\)

Complete the square: \(\ 420=\left(n^{2}+n+1 / 4\right)\)

Factor: \(\ 420=(n+1 / 2)^{2}\)

Square root both sides: \(\ 201 / 2=n+1 / 2\)

\(\ n=20\)

Express the sum using sigma notation: 1 + 3 + 9 + 27 + ...

**Solution**

\(\ \sum_{n=1}^{\infty} 3^{n-1} \text { or } \sum_{n=0}^{\infty} 3^{n}\)

Find the sum of \(\ \sum_{n=0}^{12} \frac{-11}{3}(n-1)\).

**Solution**

Let's look at both ways of solving this one:

- We could plug in all the numbers between 0 and 12 to get: \(\ \sum_{n=0}^{12} \frac{-11}{3}(n-1)\) and then add them together to get the sum.
- Using a formula: A slightly modified version of Gauss's formula looks like this: \(\ S_{n}=\frac{k}{2}\left(a_{0}+a_{n}\right)\), where
is the number of terms in the series*k*, and*plus one**a*and_{0}*a*are the first and last terms in the series_{n}

\(\ \begin{aligned}

\left(\frac{13}{2}\right)\left(\left(\frac{-11}{3}\right)(-1)+\left(\frac{-11}{3}\right)(11)\right) &=\left(\frac{13}{2}\right)\left(\frac{11}{3}+\frac{-121}{3}\right) \\

&=\left(\frac{13}{2}\right)\left(\frac{-110}{3}\right) \\

&=\frac{-1430}{6} \\

&=\frac{-715}{3} \rightarrow-238 \frac{1}{3}

\end{aligned}\)

Use Gauss's formula to find the sum of the first 200 positive integers.

**Solution**

Gauss's formula:

\(\ \begin{array}{l}

=\frac{(n)(n+1)}{2} \\

n=200 \\

=\frac{(200)(200+1)}{2} \\

\text { Sum }=20,100

\end{array}\)

## Review

Calculate the sums of the given series, you may use addition of individual terms, or a series sum formula. You may use a graphing tool for any 3 of them. Try to use each method at least once.

- \(\ \sum_{n=0}^{16}-10+3(n-1)\)
- \(\ 64+72+80+\ldots+200\)
- \(\ \frac{-15}{4}-4-\frac{17}{4}+\ldots \frac{17}{2}\)
- \(\ \sum_{n=0}^{6} 6-\frac{1}{2}(n-1)\)
- \(\ \frac{-71}{3}-\frac{67}{3}-21+\ldots+\frac{37}{3}\)
- \(\ 2+4+6+\ldots 26\)
- \(\ -2-1+0+\ldots+12\)
- \(\ -\frac{21}{4}-\frac{13}{2}-\frac{31}{4}+\ldots-\frac{71}{4}\)
- \(\ -\frac{7}{2}-8-\frac{25}{2}+\ldots-\frac{43}{2}\)
- \(\ \sum_{n=7}^{20}-1+(n-1)\)
- \(\ \sum_{n=-6}^{22} 5+3(n-1)\)
- \(\ \sum_{n=-2}^{13}-5-\frac{3}{2}(n-1)\)

Consider the sums \(\ \sum_{n=1}^{5}(n+1)\) and \(\ \sum_{n=1}^{5}(n-4)\)

- What is the product of \(\ \left(\sum_{n=1}^{5}(n+1)\right)\left(\sum_{n=1}^{5}(n-4)\right)\)
- What is the sum of \(\ \sum_{n=1}^{5}(n+1)(n-4)\)?
- Look closer at the last two problems, what does this tell you about rules for working with sums?

## Vocabulary

Term | Definition |
---|---|

Σ |
Σ (sigma) is the Greek letter meaning "the sum of" when used in mathematics. |

converge |
If a series has a limit, and the limit exists, the series converges. |

convergent |
If a series has a limit, and the limit exists, the series is convergent. |

divergent |
If a series does not have a limit, or the limit is infinity, then the series is divergent. |

diverges |
If a series does not have a limit, or the limit is infinity, then the series diverges. |

index |
The index of a sum is the variable in the sum. |

limits |
The limits of a sum are written above and below the Σ, and describe the domain to be used in the series calculation. |

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

Sigma |
Σ, pronounced syg-mah, is the Greek letter that in math means "the sum of". |

summand |
A summand is an expression being summed. It directly follows the sigma symbol. |