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Summation notation

We can describe sums with multiple terms using the sigma operator, Σ. Learn how to evaluate sums written this way.
Summation notation (or sigma notation) allows us to write a long sum in a single expression.

Unpacking the meaning of summation notation

This is the sigma symbol: sum. It tells us that we are summing something.
Let's start with a basic example:
Stop at n=3(inclusive)n=132n1Expression for eachStart at n=1term in the sum\begin{aligned} \scriptsize\text{Stop at }n=3& \\ \scriptsize\text{(inclusive)} \\ \searrow\qquad& \\\\ \LARGE\displaystyle\sum_{n=1}^3&\LARGE 2n-1 \\ &\qquad\quad\nwarrow \\ \nearrow\qquad&\qquad\scriptsize\text{Expression for each} \\ \scriptsize\text{Start at }n=1&\qquad\scriptsize\text{term in the sum} \end{aligned}
This is a summation of the expression 2, n, minus, 1 for integer values of n from 1 to 3:
=n=132n1=[2(1)1]n=1+[2(2)1]n=2+[2(3)1]n=3=1+3+5=9\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=1}^3 2\goldD n-1 \\\\ &=\underbrace{[2(\goldD 1)-1]}_{\goldD{n=1}}+\underbrace{[2(\goldD 2)-1]}_{\goldD{n=2}}+\underbrace{[2(\goldD 3)-1]}_{\goldD{n=3}} \\\\ &=1+3+5 \\\\ &=9 \end{aligned}
Notice how we substituted start color #e07d10, n, equals, 1, end color #e07d10, start color #e07d10, n, equals, 2, end color #e07d10, and start color #e07d10, n, equals, 3, end color #e07d10 into 2, start color #e07d10, n, end color #e07d10, minus, 1 and summed the resulting terms.
n is our summation index. When we evaluate a summation expression, we keep substituting different values for our index.
Problem 1
sum, start subscript, n, equals, 1, end subscript, start superscript, 4, end superscript, n, squared, equals, question mark
Choose 1 answer:

We can start and end the summation at any value of n. For example, this sum takes integer values of n from 4 to 6:
=n=46n1=(41)n=4+(51)n=5+(61)n=6=3+4+5=12\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=4}^6 \goldD n-1 \\\\ &=\underbrace{(\goldD 4-1)}_{\goldD{n=4}}+\underbrace{(\goldD 5-1)}_{\goldD{n=5}}+\underbrace{(\goldD 6-1)}_{\goldD{n=6}} \\\\ &=3+4+5 \\\\ &=12 \end{aligned}
We can use any letter we want for our index. For example, this expression has i for its index:
=i=023i5=[3(0) ⁣ ⁣5]i=0+[3(1) ⁣ ⁣5]i=1+[3(2) ⁣ ⁣5]i=2=5+(2)+1=6\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD i=0}^2 3\goldD i-5 \\\\ &=\underbrace{[3(\goldD 0)\!-\!5]}_{\goldD{i=0}}+\underbrace{[3(\goldD 1)\!-\!5]}_{\goldD{i=1}}+\underbrace{[3(\goldD 2)\!-\!5]}_{\goldD{i=2}} \\\\ &=-5+(-2)+1 \\\\ &=-6 \end{aligned}
Problem 2
sum, start subscript, k, equals, 3, end subscript, start superscript, 5, end superscript, k, left parenthesis, k, plus, 1, right parenthesis, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Problem 3
Consider the sum 4, plus, 25, plus, 64, plus, 121.
Which expression is equal to the above sum?
Choose all answers that apply:

Some summation expressions have variables other than the index. Consider this sum:
sum, start subscript, n, equals, 1, end subscript, start superscript, 4, end superscript, start fraction, k, divided by, n, plus, 1, end fraction.
Notice that our index is n, not k. This means we substitute the values into n, and k remains unknown:
=n=13kn+1=k(1)+1+k(2)+1+k(3)+1=k2+k3+k4\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=1}^3 \dfrac{k}{\goldD n+1} \\\\ &= \dfrac{k}{(\goldD 1)+1} + \dfrac{k}{(\goldD 2)+1} + \dfrac{k}{(\goldD 3)+1} \\\\ &= \dfrac{k}{2} + \dfrac{k}{3} + \dfrac{k}{4} \end{aligned}
Key takeaway: Before evaluating a sum in summation notation, always make sure you identified the index, and that you are only substituting into that index. Other unknowns should remain as they are.
Problem 4
sum, start subscript, m, equals, 1, end subscript, start superscript, 4, end superscript, 8, k, minus, 6, m, equals, question mark
Choose 1 answer:

Want more practice? Try this exercise.

Want to join the conversation?

  • piceratops ultimate style avatar for user s.v.jayachand
    can there be 2 sigma symbols at a time
    and what if such condition occurs
    (36 votes)
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    • leaf green style avatar for user kubleeka
      Yes, there can. Something like Σ³ᵢ₌₁Σ⁴ⱼ₌₁ ij

      In this case, we evaluate the innermost (rightmost) sum first. In the end, this will give us a function of i, which we then compute normally.

      The inner sum is i+2i+3i+4i, or 10i. So this simplifies to Σ³ᵢ₌₁ 10i, or 10+20+30=60.
      (106 votes)
  • duskpin tree style avatar for user G Y
    In the first section (Unpacking Sigma Notation), I've seen the index equal 0. But my calculus teacher says that the index can't be 0, because you can't have the 0th term of a sequence. But all else being equal (the sequence and summation index remaining the same), what would be the difference between a sum with i = 0 and a sum with i = 1?

    Thank you.
    (10 votes)
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    • mr pink red style avatar for user andrewp18
      Nothing really. Nothing changes if you shift all the indices down by 1. In fact, you can really start at any index you want because there's no convention that the subscript has to denote which number the term is in the sequence. Generally, people start at index 1 because it happens to be convenient to use the subscripts (and so the indices) to keep track of the number of the terms.
      (14 votes)
  • blobby green style avatar for user Grave Watcher 69421
    How do i derive the formula for summation?

    Sum from k to n i = [(n-k+1)(n+k)]/2
    (6 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      Another way to derive this formula is to let
      S = Sum from k to n of i, write this sum in two ways, add the equations, and finally divide both sides by 2. We have

      S = k + (k+1) + ... + (n-1) + n
      S = n + (n-1) + ... + (k+1) + k.

      When we add these equations, we get 2S on the left side, and n-k+1 column sums that are each n+k on the right side.

      So 2S = (n-k+1)(n+k).
      Dividing both sides by 2 gives S = [(n-k+1)(n+k)]/2.
      (6 votes)
  • orange juice squid orange style avatar for user Cassie Areff
    In my physics class the derivative of momentum was taken and the summation went from having k=1 on the bottom and N on the top to just k on the bottom, why is this? Is it the same thing, but short hand?
    (8 votes)
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  • blobby green style avatar for user Ross Metcalf
    How do I solve for the number on top of the Sigma?

    If I know the starting index, and I know the formula and the final sum, how do I solve for the ending index?

    Example, how do I solve for X?
    X
    E f(n * 500) = 18000
    n = 1
    (4 votes)
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  • blobby green style avatar for user 372447
    Why is (sigma; n=1 to 3) (n!) -n undefined?
    (3 votes)
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  • piceratops seed style avatar for user kimj171
    Additionally, does the "i" simply go up by integer values? What if I wanted a different manner of arithmetically changing "i"?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      Some books/teachers will accept writing, for example, 'step 0.5' to indicate that i is going up in increments of 1/2.
      However, this isn't necessary, since you can just change the expression inside the sum to get the same effect.

      So if you have the sum from i=1 to 10 of i², and you want to i to step by 1/3 instead of 1, you can change the expression i² to (1+(i-1)/3)², and have i go from i=1 to 28. This will get you the same sequence of numbers as just changing the step.
      (2 votes)
  • piceratops ultimate style avatar for user ZS
    What if my index is m and is in the exponent of some "not index" variable?
    (2 votes)
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  • leafers ultimate style avatar for user Nick Voelckers
    Can the index only increase by 1, or is it possible for an index to increase by a larger number?
    (2 votes)
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    • female robot grace style avatar for user loumast17
      You can decide that. you give it a starting point and ending point, but you can use terms in the function to make it go faster or slower.

      You could say the sum from i=0 to n of something with i^2. or you could even have something to the i power. But you will go through the integers with i, yes.So any manipulation of skipping numbers would have to be done in the function. For instance if you only wanted even numbers you would have a 2i in the function.
      (3 votes)
  • blobby green style avatar for user ItzakB
    Maths
    (3 votes)
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