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: $$(x - p)(x + p) = x^2 - p^2$$

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 $x$.

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.