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.

$$\begin{align} (x+2)^3 &= \binom{3}{0}x^3 + \binom{3}{1}x^2(2) + \binom{3}{2}x(2^2) + \binom{3}{3}(2^3) \\ &= (1)x^3 + (3)(2)x^2 + (3)(4)x + (1)(8) \\ &= x^3 + 6x^2 + 12x + 8 \end{align}$$

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 $$23, 27, 31, 35, \dots$$

$$\begin{align} d &= 27 - 23 = 4 \\ a_1 &= 23 \\ a_{50} &= 23 + (50-1)(4) = 23 + 49(4) = 23 + 196 = 219 \\ S_{50} &= \frac{50(23+219)}{2} = 25(242) = 6050 \end{align}$$

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 $$a_n = a_1r^{n-1}$$ 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 $$-1 < r < 1$$.

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