# A-level Mathematics/OCR/C2/Sequences and Series
## Definitions

A sequence is simply a list of numbers in a particular order. We call these numbers the terms of the sequence. For instance, 2,4,6,8 are the first four terms in the sequence of even positive integers. When we take the sum of the terms in a sequence, we get a series. For example, 2+4+6+8+... is a series.
We denote the terms in a sequence by $T_{n}$ where $n$ is the number of the term in question. For example, we have $T_{1}=2$, $T_{2}=4$, $T_{3}=6$, and so on, in the sequence described above.
A definition is a rule that tells us how to compute each term in a sequence. For example, a rule for the sequence above is $T_{n}=2n$. A relation describes how each term is related to other terms. For instance, a relation for the above sequence is $T_{n+1}=T_{n}+2$.

## Sigma ($\Sigma$) Notation

As you might have suspected, describing a series with the help of some of its terms isn't always a good idea --- if too few terms are used, the series can be ambiguous to your reader; on the other hand, you risk insulting your reader by writing out too many terms! To express a series succinctly, we use the sigma $(\Sigma )$ notation instead.
In general, a series may be written as $\sum _{k=k_{0}}^{N}f(k)$, which means "sum of all terms beginning with $f(k_{0})$ up to and including $f(N)$". Hence,
$\sum _{k=k_{0}}^{N}f(k)=f(k_{0})+f(k_{0}+1)+f(k_{0}+2)+\cdots +f(N)$.
As an example, the series 2+4+6+8+... may be written as $\sum _{k=1}^{\infty }2k$.

## Recognising Simple Progressions

A progression is just another word for a sequence. In this module, you are expected to be well-acquainted with two very common types of progressions --- the arithmetic progression and the geometric progression.
Briefly, an arithmetic progression or AP is a sequence in which each successive term is the sum of the previous term and a fixed value. An example of an AP is 1,4,7,10,..., where the difference between successive terms is 3.
A geometric progression or GP is a sequence in which each successive term is the product of the previous term and a fixed value. An example of a GP is 2,4,8,16,32,..., where each term is twice the value of the previous term.

## Arithmetic Progression (AP)

An arithmetic progression (AP) is a sequence that can be written in the following way: $a,a+d,a+2d,a+3d,a+4d,\ldots$, where $a,d$ are constants. The first term $T_{1}$ in the AP is denoted by $a$, and the common difference between subsequent terms is denoted by $d$. Thus, the series 1,4,7,10,..., is an arithmetic progression with $a=1$ and $d=3$.

### Rules

The common difference $d$ can be calculated by $d=T_{n}-T_{n-1}$, where $n=2,3,4,...$.
The $n$th term is given by $T_{n}=a+(n-1)d$.
The sum of the first $n$ terms of an AP (with $T_{1}$ as its first term and $T_{n}$ as its last term) is given by $S_{n}={\frac {n}{2}}(T_{1}+T_{n})={\frac {n}{2}}(2a+(n-1)d)$.
In fact, more generally, the sum of $n$ consecutive terms in an AP is given by ${\frac {\mathrm {number\,of\,terms} }{2}}(\mathrm {first\,term} +\mathrm {last\,term} )$.

### Example

What is the sum of the even numbers 2, 4, 6, 8, ..., 100?
The given sequence can be expressed as an AP with $T_{1}=a=2$, and $d=2$. We want the sum of the first 50 terms of the AP:
$S_{50}={\frac {50}{2}}(2+100)={\frac {50}{2}}\left(2(2)+(50-1)2\right)=2550$.

## Geometric Progression (GP)

A geometric progression (GP) is a sequence that can be written in the following way: $a,ar,ar^{2},ar^{3},ar^{4},\ldots$, where $a,r$ are constants. The first term $T_{1}$ in the GP is denoted by $a$, and the common ratio between subsequent terms is denoted by $r$.

### Rules

The common ratio $r$ can be calculated by $r={\frac {T_{n}}{T_{n-1}}}$, where $n=2,3,4,...$.
The $n$th term is given by $T_{n}=ar^{n-1}$.
The sum of the first $n$ terms of an GP is given by $S_{n}={\frac {a(1-r^{n})}{1-r}}$.
Proof of this is given by:
$S_{n}=a+ar+ar^{2}+\ldots +ar^{n-1}$
$rS_{n}=ar+ar^{2}+ar^{3}+\ldots +ar^{n}$
$S_{n}-rS_{n}=a-ar^{n}$
$S_{n}(1-r)=a(1-r^{n})$
$S_{n}={\frac {a(1-r^{n})}{1-r}}$

### Sum Of An Infinite Geometric Series

We say that the geometric series $\sum _{k=0}^{\infty }ar^{k}=a+ar+ar^{2}+ar^{3}+ar^{4}+\ldots$ is convergent if the sum to infinity $S_{\infty }$ approaches some limit. This occurs when $|r|<1$. Hence, if $|r|<1$, then
$S_{\infty }=\lim _{n\rightarrow \infty }{\frac {a(1-r^{n})}{1-r}}={\frac {a}{1-r}}$.
Proof of this is given by:
$S_{\infty }=a+ar+ar^{2}+ar^{3}+...$
$rS_{\infty }=ar+ar^{2}+ar^{3}+ar^{4}+...$
$S_{\infty }-rS_{\infty }=a$
$S_{\infty }(1-r)=a$
$S_{\infty }={\frac {a}{1-r}}$

## Binomial expressions

A binomial is a polynomial with two parts in the form $(a+b)$, such as $(x+1)$. When a binomial is raised to a power, you could simplify it by multiplying out the brackets several times. The expanded polynomial is called a binomial expansion, and all binomial expansions follow a pattern that can be used to expand binomials quicker than multiplying out several brackets. For now, we will only look at binomial expressions which are raised to positive integers.

### Expansions of $(x+1)$

Here are the expansions of $x+1$ raised to different powers.
| $(x+1)^{1}=$ | $1x+1$ |
| $(x+1)^{2}=$ | $1x^{2}+2x+1$ |
| $(x+1)^{3}=$ | $1x^{3}+3x^{2}+3x+1$ |
| $(x+1)^{4}=$ | $1x^{4}+4x^{3}+6x^{2}+4x+1$ |
| $(x+1)^{5}=$ | $1x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1$ |
If you look at the coefficient of each term, you may notice a pattern. These numbers are called binomial coefficients and are found by adding the two numbers above it.

### Pascal's triangle

Binomial coefficients are more commonly known as Pascal's triangle, named after Blaise Pascal.
The first 10 lines of Pascal's triangle are:
Since each number is found by adding the two numbers above it, it is possible to find a few lines of the triangle to help you expand binomials. For binomials raised to powers greater than 10, you should use the binomial coefficient formula.

### Binomial coefficient formula

When a binomial is raised to a large power, it may be too time consuming to find the binomial coefficients by writing out Pascal's triangle. Fortunately, there is a formula that can find any line of Pascal's triangle.
If $n$ is the power of the expansion, and $r$ is the number of the term in a single row, the binomial coefficient formula is:
${n \choose r}={\frac {n!}{r!(n-r)!}}$
The $!$ means factorial and multiplies $n$ by every integer less than itself, down to 1.
So $3!=3\times 2\times 1=6$.
To find the binomial coefficients, you use the formula with the required value of $n$, and $r=0$, $r=1$, $r=2$, and so on, until $r=n$.
Most scientific calculators will have two buttons that will be useful in this process, one is the factorial button, usually labelled n! and the other will actually find ${n \choose r}$ and is often labelled nCr or $C_{r}^{n}$. (The C stands for "choose" or "combination" which is based on the formula's use in probability.)
You should be aware that Pascal's triangle is symmetrical, so once a coefficient is repeated, you can write down the rest of the coefficients with ease.

### Expanding binomials

Now that you know how to find the coefficients in a binomial expansion, you can easily expand any binomial that is raised to a positve integer by following these simple steps:
For a binomial in the form $(a+b)^{n}$,
* Write down $a$ in descending powers, from $n$ to $0$
* Write down $b$ in ascending powers, from $0$ to $n$, making sure that you place the terms so that the powers add up to $n$
* Add the binomial coefficients to each term, either from line $n$ in Pascal's triangle (ignoring the 1 at the top), or by using the binomial coefficient formula.
You then simplify where necessary.
For example, for the expansion of $(x+2)^{4}$:
$x$ in descending powers: $x^{4}$ $+$ $x^{3}$ $+$ $x^{2}$ $+$ $x^{1}$ $+$ $x^{0}$
$2$ in ascending powers: $2^{0}$ $+$ $2^{1}$ $+$ $2^{2}$ $+$ $2^{3}$ $+$ $2^{4}$
Grouping everything together we now have:
$x^{4}$$2^{0}$ $+$ $x^{3}$$2^{1}$ $+$ $x^{2}$$2^{2}$ $+$ $x^{1}$$2^{3}$ $+$ $x^{0}$$2^{4}$
Adding in the binomial coefficients:
$1x^{4}$$2^{0}$ $+$ $4x^{3}$$2^{1}$ $+$ $6x^{2}$$2^{2}$ $+$ $4x^{1}$$2^{3}$ $+$ $1x^{0}$$2^{4}$
Finally simplifying will now give us:
$x^{4}$ $+$ $8x^{3}$ $+$ $24x^{2}$ $+$ $32x^{1}$ $+$ $16 $
This process is summarised in the equation known as the binomial theorem:
$(x+y)^{n}=\sum _{r=0}^{n}{n \choose r}x^{r}y^{n-r}$
In case you are not familiar with sigma notion this means:
$(x+y)^{n}={n \choose 0}x^{0}y^{n}+{n \choose 1}x^{1}y^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+{n \choose {n-1}}x^{n-1}y^{1}+{n \choose n}x^{n}y^{0}$
Several simplifications can be made but they aren't worth memorising as you will pick them up automatically: $(x+y)^{n}=y^{n}+nxy^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+nx^{n-1}y+x^{n}$
Dividing and Factoring Polynomials / Sequences and Series / Logarithms and Exponentials / Circles and Angles / Integration