E2.7 - Quadratic, cubic and exponential sequences | Cambridge IGCSE Mathematics

A sequence can hide a square, a cube or a repeated multiplier behind extra terms. You will use differences and comparisons with familiar sequences to uncover nth-term rules, then use those rules at any position. This lesson develops the Extended sequence methods.

General quadratic sequences

The sequence $4,13,26,43,\ldots$ is known to be quadratic. How can you find its rule when it is not simply the squares plus a fixed number?

Write $T_n$ for the term at position $n$ : the small $n$ is a subscript, not a multiplier. Thus $T_3=26$ . Subtract neighbouring terms to get first differences, then subtract neighbouring first differences to get second differences.

Position $n$	$1$	$2$	$3$	$4$
$T_n$	$4$	$13$	$26$	$43$
First differences	$9$	$13$	$17$
Second differences	$4$	$4$

A linear rule $bn+c$ contributes a constant first difference $b$ , so its second differences are zero. The squares have first differences $3,5,7,\ldots$ and second differences $2$ . Multiplying the squares by $a$ multiplies their differences by $a$ .

Quadratic sequence differences

For a quadratic sequence

T_n=an^2+bn+c\qquad (a\ne0)

the second differences equal $2a$ . Subtract $an^2$ from each term to leave a linear sequence $bn+c$ .

Not given in the List of formulas: learn the method and why it works.

For the remaining linear sequence, its common difference gives $b$ ; its first term equals $b+c$ , which gives $c$ .

Worked example 1: removing the square part

Question. Find $T_n$ and $T_8$ for the quadratic sequence $4,13,26,43,\ldots$ .

Plan. Halve the second difference, remove that multiple of the squares, then find the linear remainder.

\begin{aligned} 2a&=4\quad\Rightarrow\quad a=2 && \text{second differences}\\ \text{remainders}&=2,5,8,11 && \text{subtract }2,8,18,32\\ b&=3 && \text{remainder increases by three}\\ c&=2-3=-1 && \text{remainder at position one}\\ T_n&=2n^2+3n-1 && \text{combine both parts}\\ T_8&=128+24-1=151 && \text{substitute eight} \end{aligned}

The subtraction line compares matching positions; it does not subtract one number from the whole sequence.

Answer. $T_n=2n^2+3n-1$ , and $T_8=151$ .

Check. The rule gives $4,13,26,43$ at positions $1,2,3,4$ . Continuing differences gives the fifth term $43+21=64$ , also $2(5^2)+3(5)-1$ .

Increase every term by $5$ . This changes only the constant part of the rule; every difference stays unchanged.

Worked example 2: a shifted quadratic

Question. Find $T_n$ and $T_8$ for the quadratic sequence $9,18,31,48,\ldots$ .

Plan. Keep the same difference method, but calculate the new linear remainder.

\begin{aligned} \text{first differences}&=9,13,17 && \text{subtract neighbours}\\ \text{second differences}&=4,4\quad\Rightarrow\quad a=2 && \text{halve four}\\ T_n-2n^2&:7,10,13,16 && \text{subtract matching squares}\\ 7,10,13,16&\text{ has rule }3n+4 && \text{difference three, first term seven}\\ T_n&=2n^2+3n+4 && \text{restore the square part}\\ T_8&=128+24+4=156 && \text{substitute eight} \end{aligned}

Answer. $T_n=2n^2+3n+4$ , and $T_8=156$ .

Check. Each supplied term is $5$ above the first example, and $156-151=5$ , as the two rules predict.

Watch out

A second difference of $4$ gives $2n^2$ , not $4n^2$ . You must then find the remaining linear terms: $2n^2$ alone does not fit the given terms.

Cubic sequences and differences

A cubic rule can include a square term, a linear term and a constant as well as its cube term. One more row of differences separates it from a quadratic rule.

The cubes $1,8,27,64,125$ have first differences $7,19,37,61$ , second differences $12,18,24$ and third differences $6,6$ . A quadratic part has constant second differences, so contributes zero to the third differences.

Cubic sequence differences

For a cubic sequence

T_n=an^3+bn^2+cn+d\qquad (a\ne0)

the third differences equal $6a$ . Subtract $an^3$ , find the quadratic remainder using second differences, then find its linear remainder.

Not given in the List of formulas: learn the method and why it works.

For example, a third difference of $12$ points to $2n^3$ . Removing it reduces a cubic problem to the quadratic method you have just learned. If a remainder has no square or linear term, stop when only its constant remains.

Worked example 3: uncovering three layers

Question. A cubic sequence begins $3,19,61,141,271,\ldots$ . Find $T_n$ and $T_6$ .

Plan. Use third differences for the cube term, then strip off the square and linear parts.

\begin{aligned} \text{first differences}&=16,42,80,130 && \text{subtract terms}\\ \text{second differences}&=26,38,50 && \text{subtract again}\\ \text{third differences}&=12,12\quad\Rightarrow\quad a=2 && \text{divide by six}\\ T_n-2n^3&:1,3,7,13,21 && \text{remove the cube part}\\ \text{new second differences}&=2,2,2 && \text{square coefficient is one}\\ T_n-2n^3-n^2&:0,-1,-2,-3,-4 && \text{remove the square part}\\ T_n&=2n^3+n^2-n+1 && \text{linear remainder is }-n+1\\ T_6&=432+36-6+1=463 && \text{evaluate} \end{aligned}

Answer. $T_n=2n^3+n^2-n+1$ , and $T_6=463$ .

Check. The next second difference is $62$ , the next first difference is $192$ , and $271+192=463$ . This agrees with substitution.

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E2.7 - Quadratic, cubic and exponential sequences

General quadratic sequences

Cubic sequences and differences

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