Summation Notation

Review summation notation and some basic rules

1 Definition

• Summation is the process of adding a series or sequence of numbers.

• Its mathematical notation is \(\displaystyle \sum_{i=1}^{n}x_i\):

  • Sigma (\(\sum\)) is the summation symbol.

  • \(x_i\) is the summation term and represents the values being summed.

  • \(\displaystyle i\) is the summation index (it can also be \(j\) or \(k\),…, etc.). It is the variable that changes as the summation progresses.

  • \(i=1\) is the lower limit (starting point) of the index.

  • \(n\) is the upper limit (stopping point) of the index.

  • The lower and upper limits can be set to any value.

  • In sum,

    \[ \displaystyle \sum_{i=1}^{n}x_i =x_1 + x_2 + x_3 + .......+x_n \]

Example M.5.1

Calculate \(\displaystyle \sum_{i=1}^{4}4i-3\)

\[ \displaystyle \sum_{i=1}^{4}4i-3= \]

\[[4(1)-3] + [4(2)-3] + [4(3)-3] + [4(4)-3] =\]

\[ [4-3] + [8-3] + [12-3] + [16-3]= \]

\[ 1 + 5 + 9 + 13 = 28 \]

• This can be calculated in R using a custom function as follows:

# Option 1
sum_series1 <- function(n) {  # create a function with n as an argument
  sum(4*(1:n)-3)              # calculate the sum of the series 4i-3 from 1 to n
}
sum_series1(4)

[1] 28

# Option 2
sum_series2 <- function(n) {  # create a function with n as an argument
  sum_n <- 0                  # create an empty variable to store the sum with 0 as an initial value
    for (i in 1:n) {          # set for loop to iterate from 1 to n
    sum_n <- sum_n + 4*i-3    # add the value of 4i-3 to the sum_n variable
  }
  return(sum_n)                 # return the final value of sum_n
}

sum_series2(4)

[1] 28

Exercise M.5.1

Calculate \(\displaystyle \sum_{j=2}^{6} \frac{2j}{j-1}\)

\[ \displaystyle \sum_{j=2}^{6} \frac{2j}{j-1}= \] \[ \frac{2(2)}{2-1} + \frac{2(3)}{3-1} + \frac{2(4)}{4-1} + \frac{2(5)}{5-1} + \frac{2(6)}{6-1}= \] \[ 4 + 3 + \frac{8}{3} + \frac{10}{4} + \frac{12}{5} = \frac{437}{30} \]

Exercise M.5.2

Calculate \(\displaystyle \sum_{i=1}^{3} a-2i\)

\[ \displaystyle \sum_{i=1}^{3} a-2i = \] \[ [a-2(1)] + [a-2(2)] + [a-2(3)] = \] \[ a-2 + a-4 + a-6 = 3a-12 = 3(a-4) \]

Note

The summation index is \(i\), so its value changes as summation progresses but the value of \(a\) remains constant.

2 Common properties of summation notation

1. \(\displaystyle \sum_{i=i_0}^{n} ca_i = c \displaystyle \sum_{i=i_0}^{n} a_i\), where \(i_0\) is the lower limit, \(n\) is the upper limit, and \(c\) is a constant

2. \(\displaystyle \sum_{i=i_0}^{n} (a_i \pm b_i) = \displaystyle \sum_{i=i_0}^{n} a_i \pm \displaystyle \sum_{i=i_0}^{n} b_i\)

3. \(\displaystyle \sum_{i=i_0}^{n} a_i = \displaystyle \sum_{i=i_0}^{m} a_i + \displaystyle \sum_{i=m+1}^{n} a_i\), where \(m\) is any integer between \(i_0\) and \(n\)

4. \(\displaystyle \sum_{i=i_0}^{n} a_i = \displaystyle \sum_{i=1}^{n} a_i - \displaystyle \sum_{i=1}^{i_0-1} a_i\), where \(i_0 \ne 1\)

5. \(\displaystyle \sum_{i=1}^{n} c = cn\), where \(c\) is a constant

6. \(\displaystyle \sum_{i=1}^{n} 1 = n\)

7. \(\displaystyle \sum_{i=i_0}^{n} 1 = n - i_0 +1\), where \(i_0 \ne 1\)

8. \(\displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2}\)

9. \(\displaystyle \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\)

10. \(\displaystyle \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2\)

Note

The lower limit of summation and the term can vary for different rules, so be careful when applying them.

Warning

• \(\displaystyle \sum_{i=i_0}^{n} a_ib_i \ne \left( \displaystyle \sum_{i=i_0}^{n} a_i \right) \left( \displaystyle \sum_{i=i_0}^{n} b_i \right)\)

• \(\displaystyle \sum_{i=i_0}^{n} \frac{a_i}{b_i} \ne \frac{\displaystyle \sum_{i=i_0}^{n} a_i}{\displaystyle \sum_{i=i_0}^{n} b_i}\)

• \(\displaystyle \sum_{i=1}^{n} x_i^2 = x_1^2 + x_2^2 + .... + x_n^2 \ne \left( \displaystyle \sum_{i=1}^{n} x_i \right)^2\)

Example M.5.2

Calculate \(\displaystyle \sum_{i=1}^{6} 4i+3\)

\[ \displaystyle \sum_{i=1}^{6} 4i+3\ {\color{#0466c8}(\text {apply rule 2}\rightarrow)} = \]

\[ \left( \displaystyle \sum_{i=1}^{6} 4i \right) + \left( \displaystyle \sum_{i=1}^{6} 3 \right)\ {\color{#0466c8}(\text {apply rule 1 and 5}\rightarrow)} = \]

\[ \left( \displaystyle 4 \sum_{i=1}^{6} i \right) + (3 \times 6)\ {\color{#0466c8}(\text {apply rule 8}\rightarrow)} = \]

\[ 4 \times \frac{6(6+1)}{2} + (3 \times 6) = 102 \]

Exercise M.5.3

Calculate \(\displaystyle \sum_{j=6}^{10} j-2\)

The answer is

\[ \displaystyle \sum_{j=6}^{10} j-2 = \]

\[ \left( \displaystyle \sum_{j=6}^{10} j \right) - \left( \displaystyle \sum_{i=6}^{10} 2 \right)\ {\color{#ba181b}(\text {apply rule 4 and 1}\rightarrow)} = \] \[ \left( \displaystyle \sum_{j=1}^{10} j \right) - \left( \displaystyle \sum_{j=1}^{6-1} j \right) - \left(2 \displaystyle \sum_{i=6}^{10} 1 \right)\ {\color{#ba181b}(\text {apply rule 8 and 7}\rightarrow)} = \] \[ \frac{10(10+1)}{2} - \frac{5(5+1)}{2} - [2(10-6+1)] = \] \[ 55-15-10 = 30 \]

3 References

• Summation Notation. Columbia University. Retrieved August 16, 2024, from https://www.columbia.edu/itc/sipa/math/summation.html

• Stitz, C., & Zeager, J. Summation Notation. In Precalculus. LibreTexts. Retrieved August 16, 2024, from https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(Stitz-Zeager)/09%3A_Sequences_and_the_Binomial_Theorem/9.02%3A_Summation_Notation

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