Kuzee
Sequences

The Type aₙ₊₁ = qaₙ + d

This is the most important mixed-type recurrence relation, and you need to master the fixed point method to solve it.

The Recurrence Relation

Mixed-type recurrence relation
an+1=qan+d(q≠0,1;d≠0)a_{n+1} = qa_n + d \quad (q \neq 0, 1; d \neq 0)

Characteristic: it contains both multiplication (qanqa_n) and addition (+d+d).

Solution Method: The Fixed Point Method

Step 1: Find the Fixed Point

Let xx be the fixed point, satisfying x=qx+dx = qx + d. Solving gives:

x=d1−qx = \frac{d}{1-q}

Step 2: Construct a New Sequence

Let bn=an−xb_n = a_n - x. Then:

bn+1=an+1−x=(qan+d)−x=qan+d−d1−q=qan−qd1−q=q(an−d1−q)=q(an−x)=qbn\begin{aligned} b_{n+1} &= a_{n+1} - x \\ &= (qa_n + d) - x \\ &= qa_n + d - \frac{d}{1-q} \\ &= qa_n - \frac{qd}{1-q} \\ &= q\left(a_n - \frac{d}{1-q}\right) \\ &= q(a_n - x) \\ &= qb_n \end{aligned}

So {bn}\{b_n\} is a geometric sequence with common ratio qq.

Step 3: Find the General Term

bn=b1⋅qn−1=(a1−x)⋅qn−1b_n = b_1 \cdot q^{n-1} = (a_1 - x) \cdot q^{n-1}

Therefore:

The general term formula
an=x+(a1−x)⋅qn−1=d1−q+(a1−d1−q)⋅qn−1a_n = x + (a_1 - x) \cdot q^{n-1} = \frac{d}{1-q} + \left(a_1 - \frac{d}{1-q}\right) \cdot q^{n-1}

Worked Examples

Example 1: Basic Solution

The sequence {an}\{a_n\} satisfies a1=1a_1 = 1 and an+1=2an+3a_{n+1} = 2a_n + 3. Find the general term formula.

Solution:

Step 1: find the fixed point x=2x+3⇒x=−3x = 2x + 3 \quad \Rightarrow \quad x = -3

Step 2: construct a new sequence bn=an−(−3)=an+3b_n = a_n - (-3) = a_n + 3

Then bn+1=2bnb_{n+1} = 2b_n and b1=a1+3=4b_1 = a_1 + 3 = 4

Step 3: find the general term bn=4⋅2n−1=2n+1b_n = 4 \cdot 2^{n-1} = 2^{n+1}

an=bn−3=2n+1−3a_n = b_n - 3 = 2^{n+1} - 3

Example 2: A Practical Application

A product sold 100 units in its first year, and in each following year the sales are 1.2 times the previous year plus 50 units. Find the sales in the nnth year.

Solution:

Let the sales in the nnth year be ana_n units. Then:

  • a1=100a_1 = 100
  • an+1=1.2an+50a_{n+1} = 1.2a_n + 50

Fixed point: x=1.2x+50x = 1.2x + 50, giving x=−250x = -250

Let bn=an+250b_n = a_n + 250. Then bn+1=1.2bnb_{n+1} = 1.2b_n and b1=350b_1 = 350

bn=350⋅1.2n−1b_n = 350 \cdot 1.2^{n-1}

an=350⋅1.2n−1−250a_n = 350 \cdot 1.2^{n-1} - 250

Special Cases

When q=1q = 1

The recurrence relation becomes an+1=an+da_{n+1} = a_n + d, which is an arithmetic sequence.

When d=0d = 0

The recurrence relation becomes an+1=qana_{n+1} = qa_n, which is a geometric sequence.

Practice Problems

Exercise 1

The sequence {an}\{a_n\} satisfies a1=2a_1 = 2 and an+1=3an−4a_{n+1} = 3a_n - 4. Find the general term formula.

Reference Answer(2 个标签)
recurrence relationgeometric sequence

Solution:

Fixed point: x=3x−4x = 3x - 4, giving x=2x = 2

Let bn=an−2b_n = a_n - 2. Then bn+1=3bnb_{n+1} = 3b_n and b1=0b_1 = 0

bn=0⋅3n−1=0b_n = 0 \cdot 3^{n-1} = 0

an=bn+2=2a_n = b_n + 2 = 2

Answer: an=2a_n = 2 (a constant sequence)

Exercise 2

The sequence {an}\{a_n\} satisfies a1=5a_1 = 5 and an+1=2an+1a_{n+1} = 2a_n + 1. Find a5a_5.

Reference Answer(2 个标签)
recurrence relationgeometric sequence

Solution:

Fixed point: x=2x+1x = 2x + 1, giving x=−1x = -1

Let bn=an+1b_n = a_n + 1. Then bn+1=2bnb_{n+1} = 2b_n and b1=6b_1 = 6

b5=6⋅24=96b_5 = 6 \cdot 2^4 = 96

a5=b5−1=95a_5 = b_5 - 1 = 95

Answer: a5=95a_5 = 95

Exercise 3

The sequence {an}\{a_n\} satisfies a1=0a_1 = 0 and an+1=12an+3a_{n+1} = \frac{1}{2}a_n + 3. Find the general term formula.

Reference Answer(2 个标签)
recurrence relationgeometric sequence

Solution:

Fixed point: x=12x+3x = \frac{1}{2}x + 3, giving x=6x = 6

Let bn=an−6b_n = a_n - 6. Then bn+1=12bnb_{n+1} = \frac{1}{2}b_n and b1=−6b_1 = -6

bn=−6⋅(12)n−1=−6⋅12n−1=−122nb_n = -6 \cdot \left(\frac{1}{2}\right)^{n-1} = -6 \cdot \frac{1}{2^{n-1}} = -\frac{12}{2^n}

an=bn+6=6−122n=6−32n−2a_n = b_n + 6 = 6 - \frac{12}{2^n} = 6 - \frac{3}{2^{n-2}}

Answer: an=6−3⋅22−na_n = 6 - 3 \cdot 2^{2-n}

Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
an,an+1a_n, a_{n+1}数学符号a-sub-n / a-sub-n-plus-oneThe nnth and (n+1)(n+1)th terms of the sequence
qq数学符号qThe common ratio (a constant)
dd数学符号dThe additive term (a constant)
xx数学符号xThe fixed point
bnb_n数学符号b-sub-nThe constructed geometric sequence
a1,b1a_1, b_1数学符号a/b-sub-oneThe first terms of the original and new sequences
d1−q\frac{d}{1-q}数学符号d over one minus qThe fixed point formula

中英对照

中文术语英文术语音标说明
混合型mixed type/mɪkst taɪp/A recurrence containing both multiplication and addition
不动点fixed point/fɪkst pɔɪnt/A point satisfying x=f(x)x = f(x)
不动点法fixed point method/fɪkst pɔɪnt ˈmeθəd/The method of solving recurrence relations via fixed points
等比数列geometric sequence/ˌdʒiːəˈmetrɪk ˈsiːkwəns/A sequence with a fixed common ratio
通项公式general term formula/ˈdʒenərəl tɜːm ˈfɔːmjələ/A formula expressing the nnth term directly in terms of nn
Previous
The Type aₙ₊₁ = qaₙ
Next
The Type aₙ₊₁ = aₙ + f(n)