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Sum Notation and Properties of Sigma

Practice Sum Notation and Properties of Sigma
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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 \sum\limits_{n=1}^6 3n - 2 . How many tagged deer were reported?


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 \sum , sigma. Example: \sum\limits_{n=1}^5 n=1+2+3+4+5=15 . Beneath the sigma is the index (in this case n ) 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.

Example A

Write the terms and find the sum of the series: \sum\limits_{n=1}^6 4n+1

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

& \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

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 2^{nd} 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:


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

Example B

Write the terms and find the sum of the series: \sum\limits_{n=9}^{11} \frac{n(n+1)}{2}

Solution: Replace n with the values 9, 10 and 11 and sum the resulting series.

& \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

Using the calculator: sum(seq(x(x-1)/2,x,9,11))=136 .

More Guidance

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, i , to represent the index and n to represent the upper bound (the total number of terms) for the sum.

\sum\limits_{i=1}^n 1=n

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

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

If we let n=5 again we get \sum\limits_{i=1}^n i=1+2+3+4+5=15=\frac{5(5+1)}{2} . This one is a little harder to derive but can be illustrated using different values of n . 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.

\sum\limits_{i=1}^n i=\frac{n(n+1)(2n+1)}{6}

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

If we write the terms in the series and find their sum we get 1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55 .

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

Example C

Use one of the rules above to evaluate \sum\limits_{i=1}^{15} i^2 .

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

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

& \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

Therefore, 51 deer were reported.

Guided Practice

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

1. \sum\limits_{n=3}^7 2(n-3)

2. \sum\limits_{n=1}^7 \frac{1}{2}n+1

3. \sum\limits_{n=1}^4 3n^2-5


1. \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


2. \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


3. \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



The sum of the terms in a sequence.


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

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

Use your calculator to find the following sums.

  1. \sum\limits_{n=10}^{15} \frac{1}{2}n+3
  2. \sum\limits_{n=0}^{50} n-25
  3. \sum\limits_{n=1}^5 \left(\frac{1}{2}\right)^{n-5}
  4. \sum\limits_{n=5}^{12} \frac {n(2n+1)}{2}
  5. \sum\limits_{n-1}^{100} \frac{1}{2}n
  6. \sum\limits_{n=1}^{200} n

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

    1. \sum\limits_{n=1}^5 2n+3
    2. 3(5)+\sum\limits_{n=1}^5 2n
    1. \sum\limits_{n=1}^5 \frac{n(n+1)}{2}
    2. \frac{1}{2}\sum\limits_{n=1}^5 n(n+1)
    1. \sum\limits_{n=1}^5 4x^3
    2. 4\sum\limits_{n=1}^5 x^3
  1. Explain why each pair in questions 12-14 has the same sum.
  2. What is another way to explain the series in #11?

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