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.

#### Solve the following problems

Write the terms and find the sum of the 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)just as we did in the previous topic 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*}.

Write the terms and find the sum of the series: \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 in the chapter.

\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.

#### Solve the following problem

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 how many tagged deer 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

\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

\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

\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.