A24, A25 - Quadratic nth term and harder sequences (Higher) | AQA GCSE Mathematics

Changing differences can hide a simple algebraic rule. In A25 you will deduce quadratic nth terms and use their structure to explain properties of a whole sequence. In A24 you will work with surd ratios and other given rules, keeping exact values and checking which type of sequence you actually have.

Square terms and second differences (Higher)

The sequence $6, 12, 20, 30, \ldots$ begins with four composite numbers. Could a prime number appear later? A list cannot settle that question; a rule for every position can.

You already know

A linear sequence has a constant first difference. A quadratic sequence has a constant, non-zero second difference. You can find a linear nth term, expand brackets and keep surds exact.

Start with the square numbers:

\begin{array}{c|rrrr} n&1&2&3&4\\\hline n^2&1&4&9&16\\ \text{first differences}&&3&5&7\\ \text{second differences}&&&2&2 \end{array}

Doubling each square doubles each difference too, so $2n^2$ has second difference $4$ . Adding the same constant to every square changes none of the differences. This lets you identify the square part before dealing with the rest.

Quadratic nth term

An expression of the form $an^2+bn+c$ , where $a\ne0$ , for integer positions $n$ . The coefficient $a$ controls the second difference. For example, $2n^2+3$ is quadratic; $2n+3$ is linear, and $2^n+3$ is exponential.

Here is why the second difference is twice $a$ . The change in the square part from one position to the next is

a(n+1)^2-an^2=a(2n+1)

When $n$ increases by $1$ again, $a(2n+1)$ increases by $2a$ . The linear part $bn+c$ adds a fixed $b$ to each first difference, so it contributes zero to the second difference.

The square coefficient

For a quadratic sequence with nth term $an^2+bn+c$ ,

\text{second difference}=2a \qquad\text{so}\qquad a=\frac{\text{second difference}}{2}

Use consecutive, equally spaced positions, normally $n=1, 2, 3, \ldots$ .

Not on the formulae sheet: learn it.

The simplest quadratic rules have just a square part and a constant. Subtract the square part from each term to reveal that constant.

Worked example 1: shifted square numbers

Question. A quadratic sequence begins $4, 7, 12, 19$ . Work out its nth term. Non-calculator.

Plan. Find the square coefficient, then subtract the matching square values.

\begin{aligned} \text{first differences}&: 3, 5, 7 && \text{subtract consecutive terms} \\ \text{second difference}&=2 && \text{differences increase by }2 \\ a&=2\div2=1 && \text{use half the second difference} \\ \text{subtract }n^2&: 4-1,\ 7-4,\ 12-9,\ 19-16 && \text{compare matching positions} \\ \text{remainders}&: 3, 3, 3, 3 && \text{a fixed correction} \end{aligned}

Why this step? Subtracting $n^2$ removes the changing growth and leaves the same $3$ at every position.

Answer. $n^2+3$ .

Check. $1^2+3=4$ and $4^2+3=19$ .

Now double only the square part, keeping the constant correction at $3$ . The second difference doubles, so the coefficient must change too.

Worked example 2: twice the square part

Question. A quadratic sequence begins $5, 11, 21, 35$ . Work out its nth term. Non-calculator.

Plan. Find $a$ , then compare with $2n^2$ .

\begin{aligned} \text{first differences}&: 6, 10, 14 && \text{subtract neighbouring terms} \\ \text{second difference}&=4 && \text{each difference rises by }4 \\ a&=4\div2=2 && \text{halve, not double} \\ \text{terms of }2n^2&: 2, 8, 18, 32 && \text{positions }1, 2, 3, 4 \\ \text{remainders}&: 3, 3, 3, 3 && \text{subtract matching entries} \end{aligned}

Why this step? The second difference $4$ belongs to $2n^2$ , because the second difference of $n^2$ is already $2$ .

Answer. $2n^2+3$ .

Check. $2(1)^2+3=5$ and $2(4)^2+3=35$ .

The coefficient need not be a positive whole number. A second difference of $1$ means $a=\frac12$ ; a second difference of $-4$ means $a=-2$ . Use the signed value.

Watch out

Wrong: second difference $4$ , so start with $4n^2$ . Right: start with $2n^2$ , because $a=4\div2$ . A constant second difference identifies the square coefficient, not the whole rule.

Change the square coefficient again while keeping the same fixed correction.

Removing a linear remainder (Higher)

After removing the square part, the remainder may be linear rather than constant.

For a quadratic nth term $an^2+bn+c$ , subtracting $an^2$ leaves $bn+c$ . That remainder has no square term, so the linear method from the previous lesson finishes the job.

Finding a quadratic nth term

For a stated quadratic sequence at consecutive positions:

\begin{aligned} a&=\tfrac12\times\text{second difference}\\ \text{remainder at position }n&=\text{given term}-an^2\\ \text{nth term}&=an^2+\text{linear nth term of the remainders} \end{aligned}

Subtract term by term at matching positions. Do not subtract the same $a$ from every term.

Not on the formulae sheet: learn it.

A small table keeps all three rows aligned: the given sequence, the square part and the remainder. The remainder is a useful check on your choice of $a$ .

Worked example 3: a non-constant remainder

Question. A quadratic sequence begins $4, 11, 22, 37$ . Work out an expression for its nth term. Non-calculator.

Plan. Use the second differences to find $a$ , subtract $an^2$ , then find the linear remainder's rule.

\begin{aligned} \text{first differences}&: 7, 11, 15 && \text{next term minus previous} \\ \text{second difference}&=4 && \text{constant change} \\ a&=2 && \text{half of }4 \end{aligned}

\begin{array}{c|rrrr} n&1&2&3&4\\\hline \text{given terms}&4&11&22&37\\ 2n^2&2&8&18&32\\\hline \text{remainders}&2&3&4&5 \end{array}

\begin{aligned} \text{remainder rule}&=n+1 && \text{difference }1\text{ and first term }2 \\ \text{nth term}&=2n^2+n+1 && \text{put the pieces back together} \end{aligned}

Why this step? The remainder $2, 3, 4, 5$ still changes. Adding just its first value, $2$ , would fit only the first term.

Answer. $2n^2+n+1$ .

Check. $2(1)^2+1+1=4$ , and $2(4)^2+4+1=37$ .

Now change only the sign of the square coefficient. The second difference becomes negative, but the remainder method stays the same.

Worked example 4: a negative square coefficient

Question. A quadratic sequence begins $0, -5, -14, -27$ . Work out an expression for its nth term. Non-calculator.

Plan. Keep the signs in the difference rows, subtract the negative square part, then find the linear remainder.

\begin{aligned} \text{first differences}&: -5, -9, -13 && \text{subtract consecutive terms} \\ \text{second difference}&=-4 && -9-(-5)=-4 \\ a&=-2 && \text{half of }-4 \end{aligned}

\begin{array}{c|rrrr} n&1&2&3&4\\\hline \text{given terms}&0&-5&-14&-27\\ -2n^2&-2&-8&-18&-32\\\hline \text{remainders}&2&3&4&5 \end{array}

\begin{aligned} \text{remainder rule}&=n+1 && \text{same remainder as before} \\ \text{nth term}&=-2n^2+n+1 && \text{restore the square part} \end{aligned}

Why this step? Subtracting $-2n^2$ adds $2n^2$ . For example, the second remainder is $-5-(-8)=3$ .

Answer. $-2n^2+n+1$ .

Check. $-2(1)^2+1+1=0$ and $-2(4)^2+4+1=-27$ .

A correct subtraction leaves a linear row, whose first differences are constant. If the remainder still has changing first differences, check $a$ , the position labels and each subtraction before continuing.

Watch out

Wrong: second difference $4$ , therefore the whole rule is $2n^2$ . Right: subtract $2n^2$ and inspect what remains. The linear and constant pieces still need to be found.

Raise every term in Worked example 4 by $3$ . The square coefficient and linear coefficient stay fixed; only the constant correction changes.

Explaining properties of every term (Higher)

A factorised nth term can explain an entire sequence at once.

Return to $6, 12, 20, 30, \ldots$ . Its first differences are $6, 8, 10$ , so its second difference is $2$ and its square coefficient is $1$ . Subtracting $n^2$ gives $5, 8, 11, 14$ , whose rule is $3n+2$ . The original rule is therefore $n^2+3n+2$ .

You already know

A prime is a positive integer with exactly two positive factors. A product of two integers both greater than $1$ is composite. Factorising $n^2+3n+2$ means finding two numbers whose sum is $3$ and product is $2$ .

Checking ten terms cannot prove a claim about all positive integer positions. Instead, look for a feature of the algebra that holds at every position, such as two factors greater than one. The factorisation and prime-number facts are not on the formulae sheet: learn them.

Worked example 5: consecutive factors

Question. A sequence has nth term $n^2+3n+2$ , for positive integer $n$ . Are any terms prime? Give a reason for your answer. Non-calculator.

Plan. Factorise the expression and check the smallest possible values of both factors.

\begin{aligned} n^2+3n+2&=(n+1)(n+2) && 1+2=3\text{ and }1\times2=2 \\ n+1&\geqslant2 && n\geqslant1 \\ n+2&\geqslant3 && n\geqslant1 \end{aligned}

Why this step? The two factors are integers greater than $1$ at every allowed position, so neither can be dismissed as the trivial factor $1$ .

Answer. No. Every term is a product of two integers greater than $1$ , so every term is composite.

Check. The first term is $(1+1)(1+2)=6$ . The tenth is $11\times12=132$ , matching $10^2+3(10)+2=132$ .

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A24, A25 - Quadratic nth term and harder sequences (Higher)

Square terms and second differences (Higher)

Removing a linear remainder (Higher)

Explaining properties of every term (Higher)

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