A recurrence relation pins down a whole sequence with just a first term and one rule that makes the next term from the one before.
With a starting number and the rule, repeating the rule produces the second, third and fourth terms and so on without end.
Adding the same number every time gives an arithmetic sequence, multiplying by the same number gives a geometric one, and when the amount added changes with n, those amounts are collected with sigma to find the general term.
In the picture, raising the number of terms adds the dots one by one, starting at a₁ = 1 and adding 2 each time.
Making the next term from the one before
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The terms made by a₁ = 1, aₙ₊₁ = aₙ + 2 plotted as dots: term number across, value up
5
👀 How a recurrence defines a sequence
①A recurrence fixes a sequence with a first term and a rule that makes the next term from the previous one
②Starting at a₁ and repeating the rule fixes every term, one at a time
③Defining a sequence this way is called a recursive (inductive) definition
④The dots in the picture rise by 2 per step, so they lie on one straight line
Arithmetic and geometric types
Arithmetic-type recurrence
an+1 = an + d ⇒ arithmetic, an = a1 + (n−1)d
Adding the same value d each time gives an arithmetic sequence with common difference d
Geometric-type recurrence
an+1 = r·an ⇒ geometric, an = a1·rn-1
Multiplying by the same number r each time gives a geometric sequence with ratio r
When the amount added changes
Recurrence with varying increments
an+1 = an + f(n) ⇒ an = a1 + ∑k=1n-1 f(k)
When the amount added, f(n), changes with n, the general term is the first term plus the sum of f(1) through f(n − 1)
Worked examples
Example 1
For a1=3, an+1=an+2, find a5.
1
Each term is the previous one plus 2, so this is an arithmetic sequence with first term 3 and difference 2.
an = 3 + (n−1)·2
2
Put n = 5 into the general term.
a5 = 3 + 4·2 = 11
▸ 11
With "previous term + constant", there is no need to build the terms one by one; the arithmetic general term works straight away. Listing 3, 5, 7, 9, 11 gives the same answer.
Example 2
For a1=1, an+1=2an+1, find a4.
1
Apply the rule step by step from a₁ = 1.
a2=3, a3=7, a4=15
2
a₄ is a₃ = 7 doubled, plus 1.
a4 = 2·7 + 1 = 15
▸ 15
The form 2aₙ + 1 is neither arithmetic nor geometric, so building the first few terms is quicker. Since 1, 3, 7, 15 are 2¹ − 1, 2² − 1, 2³ − 1, 2⁴ − 1, a good guess is aₙ = 2ⁿ − 1.
Summary
Key result
an+1=an+d (arith), an+1=r an (geo), an+1=an+f(n) (diff)
Look at the form of the recurrence first and decide whether it is arithmetic, geometric or varying-increment
2020 CSAT Math type, adapted
For the sequence defined by a1=2, an+1=an+3, what is a10?
①26
②29
③32
④35
⑤38
▸ ② 29
1
Each term is the previous one plus 3, so this is an arithmetic sequence with first term 2 and difference 3.
an = 2 + (n−1)·3
2
Putting n = 10 into the general term adds 3 nine times to 2.
a10 = 2 + 9·3 = 29
🎯 Exam Points
①Identify the form first: always adding d means arithmetic, always multiplying by r means geometric
②The type aₙ₊₁ = aₙ + f(n) is solved with sigma
③If the form is unfamiliar, build a few terms from a₁ and guess the pattern
④The first term and the rule together fix the sequence completely
⑤Once the general term is found, put in the term number carefully