# 1 Number patterns and algebra

## 1.1 Arithmetic sequences

Sum of a sequence of natural numbers can be rewritten as a sum of pairs of numbers: $$1 + 2 + ... + 99 + 100 = (1 + 100) + (2 + 99) + ... + (50 + 51)$$ There are 50 pairs, each of them having a sum of $101$, so: $$1 + 2 + ... + 99 + 100 = 50 \times 101 = 5050$$

In case of a sum of successive odd numbers the sum is always the square of how many odd numbers I add, e.g. $1 + 3 + 5 + 7 = 16 = 4^2$ . For $n$ numbers the formula is $n^2$.

Formula for $n$ successive natural numbers is: $$1 + 2 + ... + n = \frac{1}{2}n(n + 1)$$ The numbers given by the formula are called triangular numbers.

Any list of numbers is called a sequence

• $1, 2, 3, ..., 100$ is finite sequence
• $1, 2, 3, ...$ is infinite sequence

The numbers in a sequence are called the terms of the sequence. Arithmetic sequence is a sequence where the difference between consequtive terms is constant, e.g.: $1, 2, 3, ...$ is a arithmetic sequence with difference of 1, To specify an arithemtic sequence we can give the first term denoted by $a$, the difference denoted by $d$. If the sequence is finite the number of terms is denoted by $n$.

The $n$th term of an arithmetic sequence with first term $a$ and difference $d$ is given by the formula $$n\text{th term} = a + (n - 1)d$$

The number of terms $n$ of a finite arithmetic sequence with first term $a$, last term $L$ and non-zero difference $d$ is given by the formula: $$n = \frac{L - a}{d} + 1$$

The sum of the finite arithmetic sequence with first term $a$, difference $d$ and numbers of term $n$ is given by the formula: $$S = \frac{1}{2}n(2a + (n - 1)d)$$ An alternative formula involving the last term $L$: $$S = \frac{1}{2}n(a + L)$$

# 2 Multiplying out pairs of brackets

## 2.1 Pairs of brackets

Strategy to multiply out two brackets

Multiply each term inside the first bracket by each term inside the second bracket, and add the resulting terms.

## 2.2 Squaring brackets

\begin{aligned} (x + p)^2 = (x + p)(x + p) \\ = x^2 + xp + px + p^2 \\ = x^2 + 2px + p^2 \end{aligned}

Hence the general formula: $$(x + p)^2 = x^2 + 2px + p^2$$ and $$(x - p)^2 = x^2 - 2px + p^2$$

## 2.3 Differences of two squares

$$(x - p)(x + p) = x^2 + xp - px - p^2$$ Hence: $$(x - p)(x + p) = x^2 - p^2$$

# 3 Quadratic expressions and equations

An expression of the form $ax^2 + bx + c$, where $a, b, c$ are numbers and $a \ne 0$ is called a quadratic expression in $x$. $a, b, c$ are called the coefficients of the quadratic.

An equation that can be expressed in the form: $$ax^2 + bx + c = 0$$ is called a quadratic equation in $x$.

## 3.3 Solving simple quadratic equations

An equation of the form $x^2 = d$, where $d > 0$ has two solutions $x = \pm \sqrt{d}$.

Equations as $x^2 + 1 = 0$ have no solution among the real numbers.

## 3.4 Factorising quadratics of the form $x^2 + bx + c$

Fill in the gaps in the brackets on the right-hand side of the equation $$x^2 + bx + c = (x...)(x...)$$ with two numbers whose product is $c$ and whose sum is $b$. I can search systematically by writing down all the factor pairs of $c$ and choosing (if possible) a pair whose sum is $b$.

## 3.5 Solving quadratic equations by factorisation

If the product of two or more numbers is 0, then at least one of the numbers must be 0. Hence in an equation as $(x - 2)(x - 3) = 0$ either $x - 2$ or $x - 3$ is 0. If $x - 2 = 0$ then $x = 2$, if $x - 3 = 0$ then $x - 3$ so the equation has two solutions.

### Strategy to solve $x^2 + bx + c = 0$ by factorisation

1. Find a factorisation: $x^2 + bx + c = (x + p)(x + p)$
2. Then $(x + p)(x + p) = 0$, so $x + p = 0$ or $x + q = 0$ and hence the solutions are $x = -p$ and $x = -q$.

When the two solutions are same it's said that equation has a repeated solution.

## 3.6 Factorising quadratics of the form $ax^2 + bx + c$

E.g. following quadratic expression; $$2x^2 - x - 6$$ Where: \begin{aligned} a &= 2\\ b &= -1\\ c &= -6 \end{aligned}

Find two numbers whose product is $ac$ and whose sum is $b$.

The numbers are $3$ and $-4$.

Rewrite the quadratic expression, splitting the term in $x$ using the above factor pair.

$$2x^2 - x - 6 = 2x^2 +3x -4x - 6$$

Group the four terms in pairs and take out common factors to give the required factorisation.

\begin{aligned} 2x^2 - x - 6 &= 2x^2 +3x -4x - 6 \\ &= x(2x + 3) -2(2x + 3) \\ &= (x - 2)(2x + 3) \end{aligned}

• If the coefficient of $x^2$ is negative, then multiply the equation through by $-1$ to make this coefficient positive.
$\frac{a}{b}$ is equivalent to $\frac{a(a + 1)}{b(b + 1)}$ and common factor can be cancelled out as usual.