Arithmetic Progression

The above progression has $a=-7$ and $d=3$. Given one term we can evaluate the next by adding the common difference $d$. We call this an arithmetic progression. The first term is $a = -7$, the second is $a+d=-4$, the third is $a+2d=-1$ and the $n^{th}$ term is $a+(n-1)d$.

Sum of an Arithmetic Progression

When the young Carl Friedrich Gauss was in primary school in the late 18th century his teacher asked the class to add the numbers $1$ to $100$. Gauss immediately walked up to the teacher with the answer written on a piece of paper.

Gauss had noticed that $1+100=101$, $2+99=101$, $3+98=101$ so the sum of numbers $1$ to $100 = 100 \times 101 \div 2=5050$.

More generally the sum to $n$ terms of an arithmetic progression with a first term $a$ and a common difference $d$ is given by
$\sum\limits_{k=0}^{k=n-1} (a+kd) = \frac{n}{2}(2a+(n-1)d)$

Geometric Progression

What is the next term in this series: $3, 6, 12, 24, 48 . . .$? In this series there is no common difference but there is a common ratio. If we divide each term by the previous term the ratios are all the same. This series, then, is a geometric progression. The first term is $a = 3$ and the common ratio is $r = 2$.

The $n^{th}$ term is given by $a r^{n-1}$

Sum of a Geometric Progression

The sum to $n$ terms of a geometric progression can be written as

$\sum\limits_{k=0}^{k=n-1} (ar^k) = a + ar^1 + ar^2 . . . + ar^{n-1}$

Multiplying both sides by $r$ we get

$r\sum\limits_{k=0}^{k=n-1} (ar^k) = ar + ar^2 + ar^3 + . . . + ar^n$

Subtracting the second equation from the first we get

$(1-r)\sum\limits_{k=0}^{k=n-1} (ar^k) = a - ar^n$

So we can write

$\sum\limits_{k=0}^{k=n-1} (ar^k) = a\frac{1-r^n}{1-r}$ or $a\frac{r^n-1}{r-1}$

Factorial Numbers

A factorial number is a natural number multiplied by each smaller natural number down to 1. Factorial numbers are written with a trailing $!$. For example, $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

$0!$ is defined to be $1$ and a number like $-9!$ means $-(9!)$ because factorials are natural numbers not integers.

Binomial Theorem

Expanding expressions like $(a+b)^2$ is easy. Expressions like $(a+b)^3$ are a bit harder. $(a+b)^4$ is harder again and you wouldn't expand $(a+b)^5$ unless there was no other option.

One other option is the binomial theorem. The binomial theorem allows us to expand two term expressions to any power, $(a+b)^n$, without sequentially expanding each bracket. The binomial theorem states:

$\begin{align} (a+b)^n = \sum\limits_{r=0}^{n} c^n_r a^{n-r} b^r \end{align}$

where $c^n_r = \frac{n!}{r!(n-r)!}$

The easiest way to expand a binomial expression is to create a table with five columns. In the first column put $r$ going from $0$ to $n$. In the second column put the $c^n_r$ terms. In the third column put the $a^{n-r}$ terms. Put $b^r$ in the fourth and the results in the fifth. As a simple example let's expand $(a+b)^2$

$c^2_0=1$, $c^2_1=2$ and $c^2_2=1$ so our table looks like this:

$r$	$c^n_r$	$a^{n-r}$	$b^r$	Result
$0$	$1$	$a^2$	$1$	$a^2$
$1$	$2$	$a$	$b$	$2ab$
$2$	$1$	$1$	$b^2$	$b^2$

Collecting the terms from the Result column we get:

$(a+b)^2=a^2+2ab+b^2$