6.06a, 6.06b - Quadratic and other sequence rules

6.06a, 6.06b - Quadratic and other sequence rules

Can the differences between terms reveal a formula for a whole sequence? In this Higher lesson, learn subscript and recurrence notation, explain the second-difference method for quadratic nth terms, and build other rules from fractions and repeated surd multipliers.

Subscripts and recurrence rules

The notation u4u_4 means the value in position 4, not u×4u\times4. Subscripts let you distinguish a position from the number stored there.

Subscript notation

unu_n is the nth term and un+1u_{n+1} is the next term. A position-to-term rule, such as un=2n+1u_n=2n+1, uses the position directly. A recurrence rule, such as un+1=3un−4u_{n+1}=3u_n-4, uses earlier term values. Give sufficient starting terms as well as a recurrence.

For un=2n+1u_n=2n+1, u4=2(4)+1=9u_4=2(4)+1=9. For un+1=3un−4u_{n+1}=3u_n-4 with u1=3u_1=3, the first three terms are 3,5,113,5,11: substitute the previous value, not its position. Unless told otherwise, nn runs through positive integers.

Worked example 1: an unknown multiplier

Question. A sequence follows un+1=kun−4u_{n+1}=ku_n-4, with u2=5u_2=5 and u3=11u_3=11. Find kk and u4u_4.

Plan. Use the known adjacent terms to find the multiplier; then move forwards once.

11=5k−415=5kk=3u4=3(11)−4=29\begin{aligned} 11&=5k-4 \\ 15&=5k \\ k&=3 \\ u_4&=3(11)-4=29 \end{aligned}

Answer. k=3k=3 and u4=29u_4=29.

Check. The change from 5 to 11 is indeed 3(5)−4=113(5)-4=11; reversing the final step gives (29+4)÷3=11(29+4)\div3=11.

Keep the recurrence with multiplier 3, but move backwards. Adding 4 reverses subtracting 4; division by 3 reverses multiplying by 3.

Worked example 2: finding the previous term

Question. A sequence follows un+1=3un−4u_{n+1}=3u_n-4. Given u4=23u_4=23, find u3u_3 and u2u_2.

Plan. Solve the recurrence for the earlier value at each step.

23=3u3−4u3=23+43=9u2=9+43=133\begin{aligned} 23&=3u_3-4 \\ u_3&=\frac{23+4}{3}=9 \\ u_2&=\frac{9+4}{3}=\frac{13}{3} \end{aligned}

Answer. u3=9u_3=9, u2=133u_2=\frac{13}{3}.

Check. 3×133−4=93\times\frac{13}{3}-4=9, then 3×9−4=233\times9-4=23. A term may be fractional even though its position is an integer.

A recurrence can stay still. With the same rule, starting at 2 gives 3(2)−4=23(2)-4=2, so every term is 2, including the 100th. This is a fixed value of the rule. By contrast, one cannot always work backwards uniquely: a multiplier of zero in un+1=0un+5u_{n+1}=0u_n+5 loses the earlier value completely.

Second differences

For a quadratic sequence, constant second differences reveal the coefficient of the squared term before you know the rest of the formula.

Start with n2n^2: its values 1,4,9,161,4,9,16 have first differences 3,5,73,5,7 and second differences 2. Multiplying the terms by aa multiplies both sets of differences by aa, giving second difference 2a2a.

Why do the other terms not change this? For un=an2+bn+cu_n=an^2+bn+c, subtract the nth term from the next:

un+1−un=a(n+1)2+b(n+1)+c−(an2+bn+c)=2an+a+b\begin{aligned} u_{n+1}-u_n &=a(n+1)^2+b(n+1)+c-(an^2+bn+c)\\ &=2an+a+b \end{aligned}

This is linear in nn; its next value is 2a2a larger. The constant cc disappears on the first subtraction and the linear contribution disappears on the second.

Second difference of a quadratic

If un=an2+bn+cu_n=an^2+bn+c, with a≠0a\ne0, then

constant second difference=2a.\text{constant second difference}=2a.

Therefore aa is half the second difference. This value applies to consecutive positions, increasing by 1; do not apply it unchanged to a table that skips positions. Not on the 2027 formula sheet: learn the reasoning.

An odd second difference gives a fractional aa. This does not prevent all the listed terms from being integers.

Worked example 3: a positive squared coefficient

Question. A quadratic sequence begins 4,11,22,37,…4,11,22,37,\ldots. Find the coefficient of n2n^2 and the next term.

Plan. Find two levels of differences, halve the second difference, then extend.

first differences:7,11,15second differences:4,4a=4÷2=2next difference=15+4=19next term=37+19=56\begin{aligned} \text{first differences}:&\quad7,11,15 \\ \text{second differences}:&\quad4,4 \\ a&=4\div2=2 \\ \text{next difference}&=15+4=19 \\ \text{next term}&=37+19=56 \end{aligned}

Answer. The coefficient is 2 and the next term is 56.

Check. The new first difference is 56−37=1956-37=19; 19−15=419-15=4, as required.

Change the sign of the second difference. Halving it must preserve that sign.

Worked example 4: a negative squared coefficient

Question. A quadratic sequence begins 20,17,10,−1,…20,17,10,-1,\ldots. Find the coefficient of n2n^2 and the next term.

Plan. Retain negative signs when subtracting consecutive terms.

first differences:−3,−7,−11second differences:−4,−4a=−4÷2=−2next difference=−11−4=−15next term=−1−15=−16\begin{aligned} \text{first differences}:&\quad-3,-7,-11 \\ \text{second differences}:&\quad-4,-4 \\ a&=-4\div2=-2 \\ \text{next difference}&=-11-4=-15 \\ \text{next term}&=-1-15=-16 \end{aligned}

Answer. The coefficient is −2-2 and the next term is −16-16.

Check. (−16)−(−1)=−15(-16)-(-1)=-15 and (−15)−(−11)=−4(-15)-(-11)=-4.

If the second difference is zero, a=0a=0, so the rule is arithmetic rather than genuinely quadratic. A finite difference table only proves a fact about the listed values. To conclude a quadratic formula governs all positions, you also need the stated quadratic class or a mathematical description of the continuing pattern.

Finding a quadratic nth term

Once you remove the squared contribution, what remains is arithmetic, so you can use a method you already know.

Continue this lesson.

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