C2.7; E2.7 - Simple quadratic and cubic sequences

C2.7; E2.7 - Simple quadratic and cubic sequences

The terms 2,5,10,17,…2,5,10,17,\ldots have gaps 3,5,7,…3,5,7,\ldots, so adding a fixed amount will not work. What familiar pattern lies just beneath them? C2.7 and E2.7 develop simple square- and cube-based nth terms, using differences to reveal their structure.

Square-based rules

Compare the opening list with squares at matching positions. Its first term is 12+1=21^2+1=2, its second is 22+1=52^2+1=5, and its third is 32+1=103^2+1=10. Every term is one more than the square of its position.

Simple quadratic sequence

A quadratic sequence has an nth term involving n2n^2 as its highest power. A simple square-based pattern can be found by comparing each term with n2n^2, for example n2+cn^2+c or a simple multiple of n2n^2.

Here nn starts at 1. Compare corresponding positions, not whichever square happens to be closest in size. An nth term must work for every given term.

Worked example 1: adding to squares

Question. Find the nth term of 2,5,10,17,26,…2,5,10,17,26,\ldots, and find its tenth term.

Plan. Compare the terms with 1,4,9,16,25,…1,4,9,16,25,\ldots, then substitute n=10n=10.

2−1=15−4=110−9=117−16=126−25=1nth term=n2+1tenth term=102+1=101\begin{aligned} 2-1&=1\\ 5-4&=1\\ 10-9&=1\\ 17-16&=1\\ 26-25&=1\\ \text{nth term}&=n^2+1\\ \text{tenth term}&=10^2+1=101 \end{aligned}

Answer. n2+1n^2+1; the tenth term is 101.

Check. The fifth term is 52+1=265^2+1=26, matching the last supplied value. The tenth term is 1 above 10×10=10010\times10=100, as the relationship requires.

Change only the constant shift. Negative early terms do not prevent a sequence from being square-based.

Worked example 2: subtracting from squares

Question. Find the nth term of −2,1,6,13,22,…-2,1,6,13,22,\ldots, and its tenth term.

Plan. Subtract the corresponding square from each term.

−2−1=−31−4=−36−9=−313−16=−322−25=−3nth term=n2−3tenth term=100−3=97\begin{aligned} -2-1&=-3\\ 1-4&=-3\\ 6-9&=-3\\ 13-16&=-3\\ 22-25&=-3\\ \text{nth term}&=n^2-3\\ \text{tenth term}&=100-3=97 \end{aligned}

Answer. n2−3n^2-3; the tenth term is 97.

Check. Substituting n=1,2,3,4,5n=1,2,3,4,5 gives exactly −2,1,6,13,22-2,1,6,13,22. Every term is 4 less than the corresponding term of the first example, including 101−4=97101-4=97.

A constant multiple can also be visible directly: 2,8,18,32,…2,8,18,32,\ldots is twice 1,4,9,16,…1,4,9,16,\ldots, so its nth term is 2n22n^2. Check the multiplication at every given position; doubling the position instead would give (2n)2=4n2(2n)^2=4n^2.

Watch out

n2+1n^2+1 means square the position and then add 1. It differs from (n+1)2(n+1)^2, which adds before squaring and starts 4,9,16,…4,9,16,\ldots.

Use the square in the same position as each supplied term.

Difference tables

The first differences of squares are not constant, but their own differences are. Arrange each new row between the terms it compares.

RowValues
Squares1,4,9,16,251, 4, 9, 16, 25
First differences3,5,7,93, 5, 7, 9
Second differences2,2,22, 2, 2

Second differences

First differences are next term minus current term. Second differences are the differences between successive first differences. A non-zero constant second difference is the characteristic difference pattern of a quadratic rule.

Why do the gaps grow by 2? The gap between consecutive squares is (n+1)2−n2=2n+1(n+1)^2-n^2=2n+1, which increases by 2 when nn increases by 1. Adding a constant to every square cancels out of the differences. Multiplying every square by a number multiplies all its differences by that number.

Worked example 3: a shifted square sequence

Question. Use differences to find the next term and a simple nth term for 3,6,11,18,27,…3,6,11,18,27,\ldots.

Plan. Find the second differences, continue the gaps, then compare with squares.

terms36111827first differences3579second differences222\begin{array}{c|rrrrr} \text{terms}&3&6&11&18&27\\ \text{first differences}&3&5&7&9&\\ \text{second differences}&2&2&2&& \end{array}

The next first difference is 9+2=119+2=11, giving 27+11=3827+11=38. Subtracting n2n^2 from each supplied term leaves 2.

Answer. The next term is 38; the nth term is n2+2n^2+2.

Check. Position 6 gives 62+2=386^2+2=38, agreeing with the difference calculation.

The next sequence doubles the square contribution, doubling its second difference. Compare with 2n22n^2, then find the constant shift.

Continue this lesson.

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