A sequence of 'numbers' or 'quantities' connected in some definite way is called a SERIES and the 'quantity' connected is called a TERM. This page covers:

- Arithmetic Progressions (AP)
- Geometric Progressions (GP)
- Sum to Infinity
- The Binomial Theorem
- Pascals Triangle

## Arithmetic Progressions (AP)

An AP is a series in which each term is formed from the proceeding term by the addition or subtraction of a constant quantity known as the common difference (d).

#### Examples

2,4,6,8,10 | d = 2 |

16,14,12,10 | d = -2 |

4,0,-4 | d = -4 |

### Equation for the n^{th} term

If we let:

- a = first term
- n = number of terms

the sequence is:

a, a+d, a+2d, a+3d...

So the n^{th} term is:

a+(n-1)d

#### Example

Find the 10^{th} term of 2, 5, 8, 11

10^{th} term = 2+(9*3)

= 29

### To find the sum to n terms of an AP

Let S_{n} = Sum of 'n' terms, then,

S_{n} = a + (a+d) + (a+2d) + (a+3d) + ...+ a+(n-1)d

Let l = last term, then,

S_{n} = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l

reversing the order of this gives:

S_{n} = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

2*S_{n} = a+l + (a+l) + (a+l) + (a+l) + ...+ (a+l) + (a+l) + (a+l) + a+l

2*S_{n} = (a+l) to n terms

2*S_{n} = (a+l)*n

S_{n} = n/2*(a+l)

but l = a+(n-1)d, so substituting this gives:

S_{n} = n/2*(2a+(n-1)d)

### Arithmetic Mean

When 3 quantities are in arithmetic progression the middle term is the arithmetic mean of the other two.

Find the arithmetic mean of a & b:

Let A be the arithmetic mean, so,

A - a = b - A

2A = a + b

A = (a+b)/2

## Geometric Progressions (GP)

A series of terms, each of which is formed by multiplying the term which proceeds it by a constant factor, known as the common ratio (r)

#### Examples

3, 6, 12, 24, 48 | r = 2 |

50, 25, 12.5, 6.25 | r = 1/2 |

+9,-27,+81 | r = -3 |

### Equation for the n^{th} term

If we let:

- a = first term
- n = number of terms

the sequence is:

a, ar, ar^{2}, ar^{3}...

So the n^{th} term is:

ar^{(n-1)}

#### Example

Find the 5^{th} term of a GP whose 4^{th} term is 5 and whose 7^{th} term is 320

First find a & r

4^{th} term: ar^{3} = 5

7^{th} term: ar^{6} = 320

dividing the 7^{th} term by the 4^{th} term gives:

r^{3} = 64

therefore:

a = 5/64

r = ^{3}√64 = 4

5^{th} term: ar^{4} = 5/64 * 4^{4} = 20

### To find the sum to n terms of an GP

Let S_{n} = Sum of 'n' terms of a series in a GP whose 1^{st}term is 'a' and common ratio is r.

S_{n} = a + ar + ar^{2} + ... + ar^{(n-2)} + ar^{(n-1)}

multiply both sides by r:

r*S_{n} = ar + ar^{2} + ar^{3}+ ... + ar^{(n-1)} + ar^{n}

subtract the last equation from the one before it:

S_{n} - r*S_{n} = a - ar^{n}

S_{n} (1- r)= a (1 - r^{n})

## Sum to Infinity

It can be shown that:

S = a/(1-r)

#### Example 1

Find the sum to infinity of: 5,-1,1/5...

a = 5

r = -1/5

S = 5/(1-(-1/5)) = 5/(6/5) = 25/6

#### Example 2

The recurring decimal 0.33333 can be written as a series:

0.3 + 0.03 + 0.003 + ...

By finding the sum to infinity express as a fraction:

S = a/(1-r) = 0.3 /(1-1/10) = 0.3/(9/10) = 3/9 = 1/3

## The Binomial Theorem

A binomial expression is a algebraic expression consisting of the sum of the two terms e.g. (a+b), (x-y), (4x^{2}+2y^{2}).

In general any such expression can be denoted by (a+b).

Expansions of (a+b):

(a+b)^{2}= a^{2}+ 2ab + b^{2}

(a+b)^{3}= a^{3}+ 3a^{2}b + 3ab^{2} + b^{3}

(a+b)^{4}= a^{4}+ 4a^{3}b + 3a^{2}b^{2}+ 4ab^{3} + b^{4}

It can be seen from the above that the coefficients follow a rule: to find the coefficient for each higher power we add the first and second to get a new second, the second and third to get a new third, the third and fourth to get a new forth and so on.

## Pascals Triangle

(a+b)^{n}= a^{n}+ n a^{(n-1)}b + n (n-1)/2! a^{(n-2)} b^{2}+ n (n-1)(n-2)/3! a^{(n-3)}b^{3} + ... + b^{n}

n | 2^{n}= |
||||||||||||||||||

0 | 1 | 1 | |||||||||||||||||

1 | 1 | 1 | 1+1 | ||||||||||||||||

2 | 1 | 2 | 1 | 1+2+1 | |||||||||||||||

3 | 1 | 3 | 3 | 1 | 1+3+3+1 | ||||||||||||||

4 | 1 | 4 | 6 | 4 | 1 | 1+4+6+4+1 | |||||||||||||

5 | 1 | 5 | 10 | 10 | 5 | 1 | 1+5+10+10+5+1 | ||||||||||||

6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | 1+6+15+20+15+6+1 | |||||||||||

7 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | 1+7+21+35+35+21+7+1 | ||||||||||

8 | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 | 1+8+28+56+70+56+28+8+1 | |||||||||

when a=1 and b=x the theorem becomes

(1+x)^{n}= 1 ^{}+ n x + n (n-1)/2! x ^{2}+ n (n-1)(n-2)/3! x ^{3} + ... + x ^{n}

#### example

Expand (2x+3)^{5}

(2x+3)^{5}= (2x)^{5}+ 5(2x)^{4}*3 + (5*4)/(1*2)(2x)^{3}(3)^{2}+(5*4*3)/(1*2*3)(2x)^{2}(3)^{3}+(5*4*3*2)/(1*2*3*4)(2x)(3)^{4}+(5*4*3*2*1)/(1*2*3*4*4)(3)^{5}

## Further Reading

For information about infinite series and there use to calculate the value of functions see this page.