The number of tagged deer reported to the game commission one Saturday is represented by the sum \begin{align*}\sum\limits_{n=1}^6 3n - 2\end{align*}. How many tagged deer were reported?

### Series and Summation Notation

A **series** is the sum of the terms in a sequence. A series is often expresses in summation notation (also called sigma notation) which uses the capital Greek letter \begin{align*}\sum\end{align*}, sigma. Example: \begin{align*}\sum\limits_{n=1}^5 n=1+2+3+4+5=15\end{align*}. Beneath the sigma is the index (in this case \begin{align*}n\end{align*}) which tells us what value to plug in first. Above the sigma is the upper limit which tells us the upper limit to plug into the rule.

Let's write the terms and find the sum of the following series.

- \begin{align*}\sum\limits_{n=1}^6 4n+1\end{align*}

Begin by replacing n with the values 1 through 6 to find the terms in the series and then add them together.

\begin{align*}& \left(4(1)-1\right)+\left(4(2)-1\right)+\left(4(3)-1\right)+\left(4(4)-1\right)+\left(4(5)-1\right)+\left(4(6)-1\right) \\ & 3+7+11+15+19+23 \\ & =78\end{align*}

**Calculator:** The graphing calculator can also be used to evaluate this sum. We will use a compound function in which we will sum a sequence. Go to **\begin{align*}2^{nd}\end{align*} STAT** (to get to the **List** menu) and arrow over to **MATH**. Select option **5: sum(** then return to the **List** menu, arrow over to **OPS** and select option **5: seq(** to get sum(seq( on your screen. Next, enter in (expression, variable, begin, end) to list the terms in a sequence. By including the sum( command, the calculator will sum the terms in the sequence for us. For this particular problem the expression and result on the calculator are:

\begin{align*}sum(seq(4x-1,x,1,6))=78\end{align*}

To obtain a list of the terms, just use \begin{align*}seq\left(4x-1,x,1,6\right)=\{3 \ \ 7 \ \ 11 \ \ 15 \ \ 19 \ \ 23\}\end{align*}

- \begin{align*}\sum\limits_{n=9}^{11} \frac{n(n+1)}{2}\end{align*}

Replace \begin{align*}n\end{align*} with the values 9, 10 and 11 and sum the resulting series.

\begin{align*}& \frac{9(9-1)}{2}+\frac{10(10-1)}{2}+\frac{11(11-1)}{2} \\ & \qquad \qquad \qquad 36+45+55 \\ & \qquad \qquad \qquad \qquad 136\end{align*}

Using the calculator: \begin{align*}sum(seq(x(x-1)/2,x,9,11))=136\end{align*}.

There are a few special series which are used in more advanced math classes, such as calculus. In these series, we will use the variable, \begin{align*}i\end{align*}, to represent the index and \begin{align*}n\end{align*} to represent the upper bound (the total number of terms) for the sum.

- \begin{align*}\sum\limits_{i=1}^n 1=n\end{align*}

Let \begin{align*}n = 5\end{align*}, now we have the series \begin{align*}\sum\limits_{i=1}^5 1=1+1+1+1+1=5\end{align*}. Basically, in the series we are adding 1 to itself \begin{align*}n\end{align*} times (or calculating \begin{align*}n\times1\end{align*}) so the resulting sum will always be \begin{align*}n\end{align*}.

- \begin{align*}\sum\limits_{i=1}^n i=\frac{n(n+1)}{2}\end{align*}

If we let \begin{align*}n=5\end{align*} again we get \begin{align*}\sum\limits_{i=1}^n i=1+2+3+4+5=15=\frac{5(5+1)}{2}\end{align*}. This one is a little harder to derive but can be illustrated using different values of \begin{align*}n\end{align*}. This rule is closely related to the rule for the sum of an arithmetic series and will be used to prove the sum formula later on.

- \begin{align*}\sum\limits_{i=1}^n i=\frac{n(n+1)(2n+1)}{6}\end{align*}

Let \begin{align*}n=5\end{align*} once more. Using the rule, the sum is \begin{align*}\frac{5(5+1)(2(5)+1)}{6}=\frac{5(6)(11)}{6}=55\end{align*}

If we write the terms in the series and find their sum we get \begin{align*}1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55\end{align*}.

The derivation of this rule is beyond the scope of this course.

Now, let's use one of the rules above to evaluate \begin{align*}\sum\limits_{i=1}^{15} i^2\end{align*}.

Using the rule \begin{align*}\sum\limits_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}\end{align*}, we get \begin{align*}\frac{15(15+1)(2(15)+1)}{6}=\frac{15(16)(31)}{6}=1240\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the number of tagged deer that were reported.

Begin by replacing n with the values 1 through 6 to find the terms in the series and then add them together.

\begin{align*}& \left(3(1)-2\right)+\left(3(2)-2\right)+\left(3(3)-2\right)+\left(3(4)-2\right)+\left(3(5)-2\right)+\left(3(6)-2\right) \\ & 1+4+7+10+13+16 \\ & =51\end{align*}

Therefore, 51 deer were reported.

**Evaluate the following. First without a calculator, then use the calculator to check your result.**

#### Example 2

Find the sum: \begin{align*}\sum\limits_{n=3}^7 2(n-3)\end{align*}.

\begin{align*}\sum\limits_{n=3}^7 2(n-3) &=2(3-3)+2(4-3)+2(5-3)+2(6-3)+2(7-3) \\
&=2(0)+2(1)+2(2)+2(3)+2(4) \\
&=0+2+4+6+8 \\
&=20\end{align*}

\begin{align*}sum(seq(2(x-3),x,3,7)=20\end{align*}

#### Example 3

Find the sum: \begin{align*}\sum\limits_{n=1}^7 \frac{1}{2}n+1\end{align*}.

\begin{align*}\sum\limits_{n=1}^7 \frac{1}{2}n+1 &=\frac{1}{2}(1)+1+\frac{1}{2}(2)+1+\frac{1}{2}(3)+1+\frac{1}{2}(4)+1+\frac{1}{2}(5)+1+\frac{1}{2}(6)+1+\frac{1}{2}(7)+1 \\
&=\frac {1}{2}+1+1+1+\frac{3}{2}+1+2+1+\frac{5}{2}+1+3+1+\frac{7}{2}+1 \\
&=\frac{16}{2}+13 \\
&=8+13 \\
&=21\end{align*}

\begin{align*}sum(seq(1/2x+1,x,1,7)=21\end{align*}

#### Example 4

Find the sum: \begin{align*}\sum\limits_{n=1}^4 3n^2-5\end{align*}.

\begin{align*}\sum\limits_{n=1}^4 3n^2-5 &=3(1)^2-5+3(2)^2-5+3(3)^2-5+3(4)^2-5 \\
&=3-5+12-5+27-5+48-5 \\
&=90-20 \\
&=70 \end{align*}

\begin{align*}sum(seq(3x^2-5,x,1,4)=70\end{align*}

### Review

Write out the terms and find the sum of the following series.

- \begin{align*}\sum\limits_{n=1}^5 2n\end{align*}
- \begin{align*}\sum\limits_{n=5}^8 n+3\end{align*}
- \begin{align*}\sum\limits_{n=10}^{15} n(n-3)\end{align*}
- \begin{align*}\sum\limits_{n=3}^7 \frac{n(n-1)}{2}\end{align*}
- \begin{align*}\sum\limits_{n=1}^6 2^{n-1}+3\end{align*}

Use your calculator to find the following sums.

- \begin{align*}\sum\limits_{n=10}^{15} \frac{1}{2}n+3\end{align*}
- \begin{align*}\sum\limits_{n=0}^{50} n-25\end{align*}
- \begin{align*}\sum\limits_{n=1}^5 \left(\frac{1}{2}\right)^{n-5}\end{align*}
- \begin{align*}\sum\limits_{n=5}^{12} \frac {n(2n+1)}{2}\end{align*}
- \begin{align*}\sum\limits_{n-1}^{100} \frac{1}{2}n\end{align*}
- \begin{align*}\sum\limits_{n=1}^{200} n\end{align*}

In problems 12-14, write out the terms in each of the series and find the sums.

- .
- \begin{align*}\sum\limits_{n=1}^5 2n+3\end{align*}
- \begin{align*}3(5)+\sum\limits_{n=1}^5 2n\end{align*}

- .
- \begin{align*}\sum\limits_{n=1}^5 \frac{n(n+1)}{2}\end{align*}
- \begin{align*}\frac{1}{2}\sum\limits_{n=1}^5 n(n+1)\end{align*}

- .
- \begin{align*}\sum\limits_{n=1}^5 4x^3\end{align*}
- \begin{align*}4\sum\limits_{n=1}^5 x^3\end{align*}

- Explain why each pair in questions 12-14 has the same sum.
- What is another way to explain the series in #11?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.4.