Repeatedly having to write out terms in a series is time consuming. Mathematicians have developed a form of notation which both shortens the process and is easy to use. It involves the use of the Greek capital letter Σ (sigma), the equivalent of the letter S, for sum.
The series 2 + 4 + 8 + ... 2^{12} can be shortened to \sum_{r=1}^{12}2^r.
This is because every term in the series is of the form 2^r, and all the values of 2^r, from r = 1 to r = 12 are added up. In this example the '2^r' is called the general term; 12 and 1 are the top and bottom limits of the sum.
Similarly, the series
60 + 60 × (0.95) + ... + 60 × (0.95)^{30}
can be abbreviated to
\sum_{r=0}^{30}60\times\left(0.95\right)^r
Often there is more than one way to use the notation. The series
\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}
has a general term that could be thought of as either \frac{r}{r + 1} or \frac{r-1}{r}.Hence the series can be written as either
\sum_{r=1}^{99}\frac{r}{r+1} or \sum_{r=2}^{100}\frac{r-1}{r}
Worked Examples
| \sum_{r=1}^9(10 - r)^2 | = (10 - 1)^2 + (10 - 2)^2 + ... + (10 - 9)^2 |
| = 9^2 + 8^2 + ... + 1^2 |
An alternative way of writing the same series is to think of it in reverse:
1^2 + 2^2 + ... + 8^2 + 9^2 = \sum_{r=1}^9 r^2
All multiples of 5 are of the form 5r, r ∈ N.
100 = 5 × 20 so the top limit is 20. The lowest multiple of 5 to be included is 5 × 1. The sum is therefore
5 + 10 + 15 + ... + 100 = \sum_{r=1}^{20}5r
Numbers ending in 3 have the form 10r + 3\ \ \ r\in\N. The first number required is 3 itself, so the bottom limit must be r = 0 . This means that the top limit must be n - 1. Hence the answer is
\sum_{r=0}^{n-1}(10r + 3)\ \left(= 3 + 13 + ... + (10n – 7)\right)
[An alternative answer is \sum_{r=1}^{n} (10r - 7)]
Exercises
