Lesson Objectives

- Learn how to evaluate a series
- Learn how to work with summation notation

## How to Evaluate a Series Using Summation Notation

In the last lesson, we learned that a sequence was a function that computes an ordered list. A series is just the sum of the terms of a sequence. Let's look at an example.

Example #1: Find the sum of the first three terms of the given sequence. $$a_n=42 - 7n$$ First, let's find a

Let's look at a few helpful summation properties/rules: $$\sum_{i=1}^{n}c=cn$$ $$\sum_{i=1}^{n}ca_i=c \sum_{i=1}^{n}a_i$$ $$\sum_{i=1}^{n}a_i + b_i=\sum_{i=1}^{n}a_i + \sum_{i=1}^{n}b_i$$ $$\sum_{i=1}^{n}a_i - b_i=\sum_{i=1}^{n}a_i - \sum_{i=1}^{n}b_i$$ $$\sum_{i=1}^{n}i=\frac{n(n + 1)}{2}$$ $$\sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ $$\sum_{i=1}^{n}i^3=\frac{n^2(n + 1)^2}{4}$$ Let's look at an example.

Example #2: Evaluate each series. $$\sum_{i=1}^{7}(5i^2 + 2)$$ Let's break up our problem using the summation properties: $$\sum_{i=1}^{7}5i^2 + \sum_{i=1}^{7}2$$ $$5\sum_{i=1}^{7}i^2 + \sum_{i=1}^{7}2$$ Let's take these on one at a time, using our formulas above: $$\sum_{i=1}^{n}c=cn$$ $$\sum_{i=1}^{7}2=14$$ Now, let's replace this and continue: $$5\sum_{i=1}^{7}i^2 + 14$$ $$\sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ $$\sum_{i=1}^{7}i^2=\frac{7(7 + 1)(2(7) + 1)}{6}$$ $$\sum_{i=1}^{7}i^2=\frac{840}{6}=140$$ Now, let's replace this and continue: $$5 \cdot 140 + 14=714$$ $$\sum_{i=1}^{7}(5i^2 + 2)=714$$

Example #1: Find the sum of the first three terms of the given sequence. $$a_n=42 - 7n$$ First, let's find a

_{1}, a_{2}, and a_{3}: $$a_1=42 - 7=35$$ $$a_2=42 - 14=28$$ $$a_3=42 - 21=21$$ Now, let's find our sum. We will use S_{3}to denote the sum of the first three terms of our sequence: $$S_3=35 + 28 + 21=84$$ When we work with series, we generally use sigma notation, which is also known as summation notation. $$\sum_{i=1}^{n}a_{i}=a_1 + a_2 + ...+a_n$$ The "i" is not the imaginary unit. This is known as the index of summation. Additionally, we have n, which is known as the upper limit of summation or the ending value. Our number 1, which "i" is set equal to, is the lower limit of summation or the starting value. Basically, you would start with plugging in a 1 for i and then a 2, and then a 3, and so on and so forth, until we get to plugging in an n as the final value. When we sum all of those values together, we get our answer.Let's look at a few helpful summation properties/rules: $$\sum_{i=1}^{n}c=cn$$ $$\sum_{i=1}^{n}ca_i=c \sum_{i=1}^{n}a_i$$ $$\sum_{i=1}^{n}a_i + b_i=\sum_{i=1}^{n}a_i + \sum_{i=1}^{n}b_i$$ $$\sum_{i=1}^{n}a_i - b_i=\sum_{i=1}^{n}a_i - \sum_{i=1}^{n}b_i$$ $$\sum_{i=1}^{n}i=\frac{n(n + 1)}{2}$$ $$\sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ $$\sum_{i=1}^{n}i^3=\frac{n^2(n + 1)^2}{4}$$ Let's look at an example.

Example #2: Evaluate each series. $$\sum_{i=1}^{7}(5i^2 + 2)$$ Let's break up our problem using the summation properties: $$\sum_{i=1}^{7}5i^2 + \sum_{i=1}^{7}2$$ $$5\sum_{i=1}^{7}i^2 + \sum_{i=1}^{7}2$$ Let's take these on one at a time, using our formulas above: $$\sum_{i=1}^{n}c=cn$$ $$\sum_{i=1}^{7}2=14$$ Now, let's replace this and continue: $$5\sum_{i=1}^{7}i^2 + 14$$ $$\sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ $$\sum_{i=1}^{7}i^2=\frac{7(7 + 1)(2(7) + 1)}{6}$$ $$\sum_{i=1}^{7}i^2=\frac{840}{6}=140$$ Now, let's replace this and continue: $$5 \cdot 140 + 14=714$$ $$\sum_{i=1}^{7}(5i^2 + 2)=714$$

#### Skills Check:

Example #1

Evaluate each series. $$\sum_{n=1}^{14}(3n - 3)$$

Please choose the best answer.

A

$$1092$$

B

$$273$$

C

$$546$$

D

$$443$$

E

$$323$$

Example #2

Evaluate each series. $$\sum_{n=1}^{15}(8n - 5)$$

Please choose the best answer.

A

$$104$$

B

$$131$$

C

$$171$$

D

$$885$$

E

$$1015$$

Example #3

Evaluate each series. $$\sum_{n=1}^{6}(5n - 7)$$

Please choose the best answer.

A

$$353$$

B

$$1027$$

C

$$504$$

D

$$63$$

E

$$126$$

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