Sigma notation

In a nutshell

Sigma, $$\Sigma$$, is a Greek capital letter, used to denote the sum of a given function between given limits. You write the lower limit below the $$\Sigma$$ and can give an upper limit above. If no upper limit is given, the sum consists of an infinite number of terms.

Limits

When using sigma notation to calculate the sum of a function, there will be an upper limit on the top and lower limit on the bottom of the notation. 

​$$\sum_{r=k}^n(r+2)$$​​

This calculation finds the sum of the terms of $$r+2$$ obtained from first substituting $$k$$ followed by each successive integer up to the final term gained by substituting $$n$$. So this sum will be:

​$$\begin{aligned}&[k+2]+[(k+1)+2]+[(k+2)+2]+...+[(n-1)+2]+[n+2]\\=&[k+2]+[k+3]+[k+4]+...+[n+1]+[n+2]\end{aligned}$$​​

Common notation

There are two common sums for which you can use sigma notation. These can be used to quickly find the sum of a given calculation.

​$$\boxed{\sum_{r=1}^n1=n \quad and \quad \sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}}$$​​

The first is the sum $$1$$s, $$n$$​ times and the second is the sum of the first $$n$$ integers. ​

Example 1

Calculate: $$\sum_{r=1}^{5}r$$.

Identify the appropriate formula:

$$\sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}$$​​

Identify the variables:

$$n=5$$​​

Substitute and calculate the sum:

$$\sum_{r=1}^{5}r = \dfrac{5(5+1)}{2} = \underline{15}$$​​

That is to say that $$1+2+3+4+5=15$$.

Finding sums of compound functions

A function involving a variable and constant can be solved by separating the terms. In the case below, $$a$$ and $$b$$ are constants.​

​$$\boxed{\sum_{r=1}^{n}(ar+b) =a\sum_{r=1}^{n}(r)+b\sum_{r=1}^{n}(1) }$$​​

Example 2

Calculate $$\sum_{r=1}^{15}(5r+3)$$.

Rewrite the calculation:

$$\sum_{r=1}^{15}(5r+3) = 5\sum_{r=1}^{15}r + 3\sum_{r=1}^{15}1$$​​

Solve each sum:

$$5\sum_{r=1}^{15}r= 5\left(\dfrac{15(15+1)}{2} \right) = 5(120) = 600$$​​

$$3\sum_{r=1}^{15}1= 3(15) = 45$$​​

Complete the calculation: 

​$$\sum_{r=1}^{15}(5r+3) = 600+45 = \underline{645}$$​​

Note: These functions in the form $$ar+b$$ represent an arithmetic sequence where $$a$$ is the common difference and the limits will provide the first and last term. Therefore, the calculations for an arithmetic series can also be used to solve the sum of these functions.

Sum of a geometric sequence

A geometric sequence will have the form $$a_k=ar^{k-1}$$ where $$r$$ is the common ratio and $$a$$ is the first term. The sum of a finite geometric sequence is given by:​

$$\boxed{\sum_{k=1}^{n}ar^{k-1} = \dfrac{a(r^n-1)}{r-1}}$$​​

Example 3

Calculate $$\sum_{k=3}^{10}(3\times 2^{k-1})$$.

This calculation has a lower limit of $$3$$. This can be adjusted as the sum of the third to tenth term will be equal to the sum of the first to tenth term, subtract the sum of the first to second term. Rewrite the calculation such that the initial $$k$$ in the sum is $$1$$. This will allow you to use the sigma formulae seen so far.$$1$$

$$\sum_{k=3}^{10}(3\times 2^{k-1}) = \sum_{k=1}^{10}(3\times 2^{k-1}) -\sum_{k=1}^{2}(3\times 2^{k-1})$$​​

Calculate the sums:

$$\sum_{k=1}^{10}(3\times 2^{k-1}) = \dfrac {3(2^{10}-1)}{2-1} = 3(2^{10}-1) = 3069$$​​

​$$\sum_{k=1}^{2}(3\times 2^{k-1}) = (3 \times 2^0)+(3\times 2^1) = 9$$​​

Complete the calculation: 

$$\sum_{k=3}^{10}(3\times 2^{k-1})= 3069 - 9 =\underline{3060}$$​​