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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.
\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:
\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

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{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:
\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

Problem 3
Consider the sum 4, plus, 25, plus, 64, plus, 121.
Which expression is equal to the above sum?
\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}