If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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.
This is a summation of the expression $2n-1$ for integer values of $n$ from $1$ to $3$:
$\begin{array}{rl}& \phantom{=}\sum _{n=1}^{3}2n-1\\ \\ & =\underset{n=1}{\underset{⏟}{\left[2\left(1\right)-1\right]}}+\underset{n=2}{\underset{⏟}{\left[2\left(2\right)-1\right]}}+\underset{n=3}{\underset{⏟}{\left[2\left(3\right)-1\right]}}\\ \\ & =1+3+5\\ \\ & =9\end{array}$
Notice how we substituted $n=1$, $n=2$, and $n=3$ into $2n-1$ 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
$\sum _{n=1}^{4}{n}^{2}=?$

We can start and end the summation at any value of $n$. For example, this sum takes integer values of $n$ from $4$ to $6$:
$\begin{array}{rl}& \phantom{=}\sum _{n=4}^{6}n-1\\ \\ & =\underset{n=4}{\underset{⏟}{\left(4-1\right)}}+\underset{n=5}{\underset{⏟}{\left(5-1\right)}}+\underset{n=6}{\underset{⏟}{\left(6-1\right)}}\\ \\ & =3+4+5\\ \\ & =12\end{array}$
We can use any letter we want for our index. For example, this expression has $i$ for its index:
$\begin{array}{rl}& \phantom{=}\sum _{i=0}^{2}3i-5\\ \\ & =\underset{i=0}{\underset{⏟}{\left[3\left(0\right)\phantom{\rule{-0.167em}{0ex}}-\phantom{\rule{-0.167em}{0ex}}5\right]}}+\underset{i=1}{\underset{⏟}{\left[3\left(1\right)\phantom{\rule{-0.167em}{0ex}}-\phantom{\rule{-0.167em}{0ex}}5\right]}}+\underset{i=2}{\underset{⏟}{\left[3\left(2\right)\phantom{\rule{-0.167em}{0ex}}-\phantom{\rule{-0.167em}{0ex}}5\right]}}\\ \\ & =-5+\left(-2\right)+1\\ \\ & =-6\end{array}$
Problem 2
$\sum _{k=3}^{5}k\left(k+1\right)=$

Problem 3
Consider the sum $4+25+64+121$.
Which expression is equal to the above sum?

Some summation expressions have variables other than the index. Consider this sum:
$\sum _{n=1}^{4}\frac{k}{n+1}$.
Notice that our index is $n$, not $k$. This means we substitute the values into $n$, and $k$ remains unknown:
$\begin{array}{rl}& \phantom{=}\sum _{n=1}^{3}\frac{k}{n+1}\\ \\ & =\frac{k}{\left(1\right)+1}+\frac{k}{\left(2\right)+1}+\frac{k}{\left(3\right)+1}\\ \\ & =\frac{k}{2}+\frac{k}{3}+\frac{k}{4}\end{array}$
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
$\sum _{m=1}^{4}8k-6m=\phantom{\rule{0.167em}{0ex}}?$

Want more practice? Try this exercise.

## Want to join the conversation?

• can there be 2 sigma symbols at a time
and what if such condition occurs
• 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.
• 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.
• 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.
• How do i derive the formula for summation?

Sum from k to n i = [(n-k+1)(n+k)]/2
• 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.
• 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?
• 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
• Why is (sigma; n=1 to 3) (n!) -n undefined?
• I have no idea, it doesn't look like it should be undefined. If we just calculate the sum from 1 to 3, we get a perfectly defined number:
Sum = (1! - 1) + (2! - 2) + (3! - 3)
= 0 + 0 + 3 = 3
If you set n to infinity though, the series will diverge and there will be no sum. Could you be confusing it with that?
• What would a function like this look like in Khan Academy Javascript?
• Basic summation demo with example in article:

function f(x){ return 2*x - 1; }

// Summation
function summation(lower, upper, f){
var sum = 0;
for(var i = lower; i <= upper; i++){
sum += f(i);
}
return sum;
}

// Result gives 9, which is equal to what's expected.
// Arguments: lower bound on x, upper bound on x, f(x)
println(summation(1, 3, f));

You can modify the function 'f' in any way you want.