What you'll learn
- How to interpret and use a formula for the nth term of a sequence.
- How to generate a sequence from a recurrence relation such as xn+1=f(xn)x_{n+1}=f(x_n)xn+1=f(xn).
- How to distinguish direct nth-term formulae from recursive definitions.
- How to investigate the long-term behaviour of recursively generated sequences.
What is a sequence?
A sequence is an ordered list of terms. Each term has a position, or index, within the list.
For example,
3, 7, 11, 15,…3,\ 7,\ 11,\ 15,\ldots3, 7, 11, 15,…is a sequence whose first term is 3, second term is 7, and so on. The order matters: rearranging the terms usually produces a different sequence.
Sequence notation
The nth term of a sequence is commonly written as unu_nun or xnx_nxn. Here, the subscript nnn gives the position of the term, so u1u_1u1 is the first term and u10u_{10}u10 is the tenth term.
A sequence is different from a series. A sequence is a list of terms, whereas a series is formed by adding terms of a sequence.
Sequence or series?
Do not use the words interchangeably. For example, 1,2,3,…1,2,3,\ldots1,2,3,… is a sequence, while 1+2+3+⋯1+2+3+\cdots1+2+3+⋯ is a series.
Sequences defined by an nth-term formula
An nth-term formula gives a term directly from its position. It is sometimes called an explicit formula because you can calculate any term without finding the preceding terms.
For example, if
un=3n+2,u_n=3n+2,un=3n+2,then you find a particular term by substituting its index for nnn.
Generating terms from an nth-term formula
The sequence is defined by un=n2−2n+4u_n=n^2-2n+4un=n2−2n+4. Find the first four terms and the tenth term.
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Substitute n=1,2,3,4n=1,2,3,4n=1,2,3,4 into the formula:
u1=12−2(1)+4=3,u2=22−2(2)+4=4,u3=32−2(3)+4=7,u4=42−2(4)+4=12.\begin{aligned} u_1&=1^2-2(1)+4=3,\\ u_2&=2^2-2(2)+4=4,\\ u_3&=3^2-2(3)+4=7,\\ u_4&=4^2-2(4)+4=12. \end{aligned}u1u2u3u4=12−2(1)+4=3,=22−2(2)+4=4,=32−2(3)+4=7,=42−2(4)+4=12. -
Therefore, the first four terms are
3, 4, 7, 12.3,\ 4,\ 7,\ 12.3, 4, 7, 12. -
Substitute n=10n=10n=10 directly to find the tenth term:
u10=102−2(10)+4=84.u_{10}=10^2-2(10)+4=84.u10=102−2(10)+4=84.
Check the starting index
Most OCR questions begin with n=1n=1n=1, but this is not automatic. A sequence may begin with n=0n=0n=0 or another value, so check the definition carefully.
Finding whether a value occurs in a sequence
To test whether a number is a term of a sequence, set the nth-term formula equal to that number and solve for nnn.
The solution must be an allowed integer index. A positive non-integer solution does not identify a position in the sequence.
Testing membership of a sequence
A sequence has nth term un=n2+3nu_n=n^2+3nun=n2+3n. Determine whether 70 is a term of the sequence.
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Set the formula equal to 70:
n2+3n=70.n^2+3n=70.n2+3n=70. -
Rearrange and factorise:
n2+3n−70=0,(n+10)(n−7)=0.\begin{aligned} n^2+3n-70&=0,\\ (n+10)(n-7)&=0. \end{aligned}n2+3n−70(n+10)(n−7)=0,=0. -
The solutions are n=−10n=-10n=−10 and n=7n=7n=7. Since the sequence is indexed by positive integers, only n=7n=7n=7 is valid. Therefore, 70 is the seventh term.
Accepting an invalid index
After solving for nnn, check that it is an integer in the stated domain. A negative, fractional or otherwise excluded value of nnn does not correspond to a term.
Sequences defined recursively
A sequence can also be generated from earlier terms. A rule that defines a term using one or more previous terms is called a recurrence relation or recursive formula.
The OCR specification includes simple relations of the form
xn+1=f(xn).x_{n+1}=f(x_n).xn+1=f(xn).This means: apply the function fff to the current term xnx_nxn to produce the next term xn+1x_{n+1}xn+1.
Initial value
An initial value, such as x1=3x_1=3x1=3, gives the term from which the recurrence begins. Both the recurrence relation and an initial value are normally needed to determine a unique sequence.
For instance, xn+1=2xn+1x_{n+1}=2x_n+1xn+1=2xn+1 is not enough to determine a single sequence. Different choices of x1x_1x1 produce different terms.
Work one term at a time
An nth-term formula lets you jump directly to any term. A recurrence relation usually requires you to calculate each preceding term in order.
Generating a recursive sequence
A sequence is defined by x1=3x_1=3x1=3 and
xn+1=2xn−5.x_{n+1}=2x_n-5.xn+1=2xn−5.Find x2x_2x2, x3x_3x3 and x4x_4x4.
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Use n=1n=1n=1, so that xn+1=x2x_{n+1}=x_2xn+1=x2 and xn=x1x_n=x_1xn=x1:
x2=2x1−5=2(3)−5=1.x_2=2x_1-5=2(3)-5=1.x2=2x1−5=2(3)−5=1. -
Use the newly calculated value to find the next term:
x3=2x2−5=2(1)−5=−3.x_3=2x_2-5=2(1)-5=-3.x3=2x2−5=2(1)−5=−3. -
Repeat the recurrence once more:
x4=2x3−5=2(−3)−5=−11.x_4=2x_3-5=2(-3)-5=-11.x4=2x3−5=2(−3)−5=−11. -
The sequence begins
3, 1, −3, −11,…3,\ 1,\ -3,\ -11,\ldots3, 1, −3, −11,…
Substituting the index instead of the term
In xn+1=f(xn)x_{n+1}=f(x_n)xn+1=f(xn), the input to fff is the value of the current term, not the number nnn. For example, if x2=5x_2=5x2=5, then x3=f(5)x_3=f(5)x3=f(5).
Non-linear recurrence relations
The function fff does not have to be linear. It may involve powers, fractions, roots or other functions.
With a non-linear recurrence, brackets are especially important because the whole previous term must be substituted into the rule.
Using a non-linear recurrence
A sequence is defined by x1=2x_1=2x1=2 and
xn+1=xn2−3.x_{n+1}=x_n^2-3.xn+1=xn2−3.Find the next three terms.
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Substitute x1=2x_1=2x1=2:
x2=22−3=1.x_2=2^2-3=1.x2=22−3=1. -
Use x2=1x_2=1x2=1 as the next input:
x3=12−3=−2.x_3=1^2-3=-2.x3=12−3=−2. -
Square the whole value x3=−2x_3=-2x3=−2:
x4=(−2)2−3=1.x_4=(-2)^2-3=1.x4=(−2)2−3=1. -
The values now alternate between 1 and −2-2−2, so the sequence begins
2, 1, −2, 1, −2,…2,\ 1,\ -2,\ 1,\ -2,\ldots2, 1, −2, 1, −2,…
Use brackets for negative terms
When substituting a negative term into a power, write it in brackets. In the example above, (−2)2=4(-2)^2=4(−2)2=4, whereas −22=−4-2^2=-4−22=−4 by the usual order of operations.
Long-term behaviour
A sequence converges if its terms approach a finite value as nnn becomes very large. This value is called the limit.
If the terms do not approach a finite value, the sequence diverges. A divergent sequence might grow without bound, decrease without bound, or oscillate without settling.
For a recurrence xn+1=f(xn)x_{n+1}=f(x_n)xn+1=f(xn), a possible limit LLL must satisfy
L=f(L).L=f(L).L=f(L).This is because, if both xnx_nxn and xn+1x_{n+1}xn+1 approach the same limit, replacing each by LLL gives this equation.
Finding a possible limit
A sequence is defined by x1=8x_1=8x1=8 and
xn+1=xn+63.x_{n+1}=\frac{x_n+6}{3}.xn+1=3xn+6.Find the first few terms and the possible limit.
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Generate terms using the recurrence:
x2=8+63=143,x3=143+63=329,x4=329+63=8627.\begin{aligned} x_2&=\frac{8+6}{3}=\frac{14}{3},\\ x_3&=\frac{\frac{14}{3}+6}{3}=\frac{32}{9},\\ x_4&=\frac{\frac{32}{9}+6}{3}=\frac{86}{27}. \end{aligned}x2x3x4=38+6=314,=3314+6=932,=3932+6=2786.These values move towards 3.
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If the sequence converges to LLL, then both consecutive terms approach LLL. Set
L=L+63.L=\frac{L+6}{3}.L=3L+6. -
Solve the equation:
3L=L+6,2L=6,L=3.\begin{aligned} 3L&=L+6,\\ 2L&=6,\\ L&=3. \end{aligned}3L2LL=L+6,=6,=3.Therefore, 3 is the possible limit.
A fixed point is not proof of convergence
Solving L=f(L)L=f(L)L=f(L) finds possible limits, also called fixed points. It does not by itself prove that the sequence converges: the starting value and the behaviour of the recurrence still matter.
A recurrence that does not approach its fixed point
A sequence is defined by x1=2x_1=2x1=2 and xn+1=4−xnx_{n+1}=4-x_nxn+1=4−xn. Investigate its behaviour.
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Generate several terms:
x2=2,x3=2.x_2=2,\qquad x_3=2.x2=2,x3=2.Here the initial value is already fixed, so the sequence remains constant.
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Change the initial value to x1=1x_1=1x1=1. The terms become
1, 3, 1, 3,…1,\ 3,\ 1,\ 3,\ldots1, 3, 1, 3,…so this sequence oscillates and does not converge.
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Solving the fixed-point equation gives
L=4−L⇒L=2.L=4-L\Rightarrow L=2.L=4−L⇒L=2.Although 2 is a fixed point, starting at 1 does not produce a sequence approaching it. This confirms that the fixed-point equation alone is insufficient.
Choosing the right representation
An nth-term formula is convenient when you need a distant term, because you can substitute its index directly. A recurrence relation is useful when a process naturally develops from one stage to the next, such as repeated growth, depreciation or iteration.
In modelling questions, interpret the index carefully. For example, x0x_0x0 might represent an initial population and x1x_1x1 the population after one year.
In the exam
- Identify whether you have an nth-term formula or a recurrence relation, and check the starting index.
- For a recurrence, write each new term before using it to calculate the next; keep exact fractions unless a decimal is requested.
- When finding a possible limit, state that you are assuming convergence, set xn+1x_{n+1}xn+1 and xnx_nxn equal to LLL, then solve L=f(L)L=f(L)L=f(L).
- Check indices, signs and brackets, particularly when negative terms are raised to powers.
Check yourself
- How would you determine whether 150 occurs in the sequence un=2n2−3nu_n=2n^2-3nun=2n2−3n?
- Given x1=4x_1=4x1=4 and xn+1=7−12xnx_{n+1}=7-\frac{1}{2}x_nxn+1=7−21xn, can you generate the next three terms?
- Why does solving L=f(L)L=f(L)L=f(L) not guarantee that every sequence generated by xn+1=f(xn)x_{n+1}=f(x_n)xn+1=f(xn) converges to LLL?
