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Sum Notation and Properties of Sigma

Identify and state the sum of terms in finite series

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Series and Summation Notation

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.

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

Use your calculator to find the following sums.

  1. \begin{align*}\sum\limits_{n=10}^{15} \frac{1}{2}n+3\end{align*}
  2. \begin{align*}\sum\limits_{n=0}^{50} n-25\end{align*}
  3. \begin{align*}\sum\limits_{n=1}^5 \left(\frac{1}{2}\right)^{n-5}\end{align*}
  4. \begin{align*}\sum\limits_{n=5}^{12} \frac {n(2n+1)}{2}\end{align*}
  5. \begin{align*}\sum\limits_{n-1}^{100} \frac{1}{2}n\end{align*}
  6. \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.

  1. .
    1. \begin{align*}\sum\limits_{n=1}^5 2n+3\end{align*}
    2. \begin{align*}3(5)+\sum\limits_{n=1}^5 2n\end{align*}
  2. .
    1. \begin{align*}\sum\limits_{n=1}^5 \frac{n(n+1)}{2}\end{align*}
    2. \begin{align*}\frac{1}{2}\sum\limits_{n=1}^5 n(n+1)\end{align*}
  3. .
    1. \begin{align*}\sum\limits_{n=1}^5 4x^3\end{align*}
    2. \begin{align*}4\sum\limits_{n=1}^5 x^3\end{align*}
  4. Explain why each pair in questions 12-14 has the same sum.
  5. 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. 

Vocabulary

arithmetic series

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

geometric series

A geometric series is a geometric sequence written as an uncalculated sum of terms.

sequence

A sequence is an ordered list of numbers or objects.

series

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

sigma notation

Sigma notation is also known as summation notation and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation.

summation

Sigma notation is also known as summation notation and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation.

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