Imagine you are given this set of numbers: $-7, -4, -1, 2, 5$ you can probably see what the next number will be. You can see there is a relationship between a number and the next. We call this set of numbers a **series** or a **progression**. If we subtract any number, after the first, from the previous number we can see there is a common difference of 3 which means we can define this series by the first number $a=-7$ and the common difference $d=3$. Numbers in a series are called **terms**.

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 $a+(n-1)d$.

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

$\sum\limits_{k=0}^{k=n-1} (a+kd) = \frac{n}{2}(2a+(n-1)d)$

**Example 12.1**: Find the sum to 10 terms of the following series:

$3, 7, 11, 15, 19 . . .$

The second term minus the first term is $7-3=4$. The third term minus the second term is $11-7=4$ so the series is arithmetic and the common difference $d=4$.

The sum to 10 terms is given by:

\begin{align} sum_{10} & = \frac{10}{2}(2 \times 3+(10-1) \times 4) \\\\ & = 5(6+36) \\\\ & = 210 \end{align}

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 $a r^{n-1}$

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 $a\frac{r^n-1}{r-1}$

**Example 12.2**: Find the sum to 10 terms of the following series:

$3, 6, 12, 24, 48 . . .$

The second term minus the first term is $6-3=3$. The third term minus the second term is $12-6=6$ so the series is not arithmetic. The second term divided by the first term is $6/3=2$. The third term divided by the second term is $12/6=2$ so the series is geometric and the common ratio $r=2$.

The sum to 10 terms is given by:

\begin{align} sum_{10} & = 3 \times \frac{2^{10}-1}{2-1} \\\\ & = 3 \times \frac{1024-1}{2-1} \\\\ & = 3069 \end{align}

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.

**Example 12.3**: Evaluate the following factorials $3!, 6!, 9!, -4!, 0!$

$3!=3 \times 2 \times 1 = 6$

$6! = 720$

$9!=362880$

$-4!=-(4!)=-24$

$0!=1$

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:

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

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

The easiest way to expand a binomial expression is to create a table with four columns. In the first column put the $c^n_r$ terms, in the second put the $a^{n-r}$ terms, $b^r$ in the third and the result in the fourth. As an 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:

$r$ | $n^c_r$ | $a^{n-r}$ | $b^r$ | 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:

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

**Example 12.4**: Expand $(x+a)^3$

$c^3_0=1$, $c^3_1=3$, $c^3_2=3$ and $c^3_3=1$ so our table looks like this:

$r$ | $n^c_r$ | $a^{n-r}$ | $b^r$ | Result |

$0$ | $1$ | $a^3$ | $1$ | $a^3$ |

$1$ | $3$ | $a^2$ | $b$ | $3a^2b$ |

$2$ | $3$ | $a$ | $b^2$ | $3ab^2$ |

$3$ | $1$ | $1$ | $b^3$ | $b^3$ |

So $(x+a)^3=a^3+3a^2b+3ab^2+b^3$

**Example 12.5**: Expand $(x+y)^6$

$c^6_0=1$, $c^6_1=6$, $c^6_2=15$, $c^6_3=20$, $c^6_4=15$, $c^6_5=6$ and $c^6_6=1$ so our table looks like this:

$r$ | $n^c_r$ | $a^{n-r}$ | $b^r$ | Result |

$0$ | $1$ | $x^6$ | $1$ | $x^6$ |

$1$ | $6$ | $x^5$ | $y$ | $6x^5y$ |

$2$ | $15$ | $x^4$ | $y^2$ | $15x^4y^2$ |

$3$ | $20$ | $x^3$ | $y^3$ | $20x^3y^3$ |

$4$ | $15$ | $x^2$ | $y^4$ | $15x^2y^4$ |

$5$ | $6$ | $x$ | $y^5$ | $6xy^5$ |

$6$ | $1$ | $1$ | $y^6$ | $y^6$ |

So $(x+y)^6=$

$x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$