# A-level Mathematics/CIE/Pure Mathematics 1/Series

## The Binomial Theorem

Before we discuss the binomial theorem, we need to discuss combinations. In order to discuss combinations, we need to discuss factorials.

### Factorials

The factorial of a number is the product of all numbers from 1 to that number. It is represented by the symbol $!$ after the number.

e.g. $4!=4\times 3\times 2\times 1=24$ The factorial can be formally defined as:

$n!={\begin{cases}1,&{\text{if }}n=0\\n\times (n-1)!,&{\text{otherwise}}\end{cases}}$ ### Combinations

Combinations are a way of calculating how many ways a set of items with a given size can be selected from a larger set of items. It is typically represented either by the column notation ${\binom {n}{k}}$ or by the notation $nCk$ .

Combinations can be calculated using factorials: ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}},n\geq k$ .

e.g. ${\binom {5}{3}}={\frac {5!}{3!2!}}={\frac {120}{6(2)}}={\frac {120}{12}}=10$ ### The Binomial Theorem

The binomial theorem is used when we need to raise a binomial, an expression consisting of two terms, to the power of a given $n$ , e.g. $(x+2)^{3}$ .

The binomial theorem states that $(a+b)^{n}={\binom {n}{0}}a^{n}+{\binom {n}{1}}a^{n-1}b+{\binom {n}{2}}a^{n-2}b^{2}+\dots +{\binom {n}{n}}b^{n}$ e.g.

{\begin{aligned}(x+2)^{3}&={\binom {3}{0}}x^{3}+{\binom {3}{1}}x^{2}(2)+{\binom {3}{2}}x(2^{2})+{\binom {3}{3}}(2^{3})\\&=(1)x^{3}+(3)(2)x^{2}+(3)(4)x+(1)(8)\\&=x^{3}+6x^{2}+12x+8\end{aligned}} The binomial theorem is sometimes summarised as $(a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b^{k}$ ## Arithmetic Progressions

An arithmetic sequence is a progression in which the numbers increment by a fixed quantity from one term to the next.

e.g. $6,8,10,12,14,\dots$ is an arithmetic sequence (the fixed quantity is $2$ )

### The nth term

The nth term of an arithmetic sequence can be determined using $a_{n}=a_{1}+(n-1)d$ where $a_{n}$ is the nth term, $a_{1}$ is the first term, and $d$ is the difference between two consecutive terms in the progression.

e.g. The sequence $4,7,10,13,\dots$ has a difference of $7-4=3$ . So the nth term of this sequence can be determined by $a_{n}=4+3(n-1)=4+3n-3=1+3n$ . Thus, if we wanted to find the 1000th term of the progression, we can use the nth term formula: $a_{1000}=1+3(1000)=3001$ .

### Sum of the first n terms

The sum of the first n terms of an arithmetic progression can be found using the formula: $S_{n}={\frac {n(a_{1}+a_{n})}{2}}$ e.g. Find the sum of the first 50 terms of the sequence $23,27,31,35,\dots$ {\begin{aligned}d&=27-23=4\\a_{1}&=23\\a_{50}&=23+(50-1)(4)=23+49(4)=23+196=219\\S_{50}&={\frac {50(23+219)}{2}}=25(242)=6050\end{aligned}} ## Geometric Progressions

A geometric progression is like an arithmetic progression except that instead of adding a constant from one term to the next, we multiply each term by a constant to get the next term.

e.g. $3,6,12,24,48,\dots$ is a geometric sequence.

### The nth term

The nth term for a geometric progression is given by $a_{n}=a_{1}r^{n-1}$ where $a_{n}$ is the nth term, $a_{1}$ is the first term, and $r$ is the ratio between two consecutive terms.

### Sum of the first n terms

The sum of the first n terms of a geometric series can be found using ${\frac {a(1-r^{n})}{1-r}}$ .

e.g. The sum of the first 10 terms of the sequence $3,6,12,24,48,\dots$ is ${\frac {3(1-2^{10})}{1-2}}={\frac {3(-1023)}{-1}}=3(1023)=3069$ .

### Convergence

A convergent geometric progression is one where the terms get smaller and smaller, meaning that as $n$ approaches infinity, the $n$ th term approaches zero. An important consequence of this is that the progression will have a defined sum to infinity.

We can tell if a sequence is convergent if the ratio $r$ is less than $1$ and more than $-1$ . If this condition is not satisfied, the sequence is divergent.

### Sum to Infinity

The sum to infinity of a geometric progression is the value that the sum of the first $n$ terms as $n$ approaches infinity. If a progression is convergent, its sum to infinity will be finite.

The sum to infinity is given by $S_{\infty }={\frac {a(1-r^{\infty })}{1-r}}$ which is equivalent to $S_{\infty }={\frac {a}{1-r}}$ if $-1 .

e.g. The sum to infinity of the sequence $1,{\frac {1}{2}},{\frac {1}{4}},{\frac {1}{8}},\dots$ is ${\frac {1}{1-{\tfrac {1}{2}}}}={\frac {1}{\tfrac {1}{2}}}=2$ 