7.3 Recurrence Relations and General Term Determination

7.3 Recurrence Relations and General Term Determination

Determining the general term $a_n$ from a recurrence relation is a central topic in Gaokao Mathematics. This section provides a comprehensive classification of standard recurrence structures and their exact resolution algorithms.

1. Difference Accumulation Method (累加法)

Applicable Type

Recurrences of the form:

$$a_{n+1} - a_n = f(n)$$

where $f(n)$ is a solvable explicit function of $n$ (such as a polynomial, exponential, or rational expression whose sum $\sum f(k)$ can be computed).

Algorithm

Write out the equations for successive indices up to $n-1$:

$$\begin{aligned} a_2 - a_1 \&= f(1) \\ a_3 - a_2 \&= f(2) \\ \&\vdots \\ a_n - a_{n-1} \&= f(n-1) \end{aligned}$$

Summing all $n-1$ equations produces a telescoping sum on the left:

$$(a_n - a_{n-1}) + \dots + (a_2 - a_1) = \sum_{k=1}^{n-1} f(k) \implies a_n = a_1 + \sum_{k=1}^{n-1} f(k) \quad (n \ge 2)$$

Mandatory Check: Always verify whether the derived formula holds for $n=1$.

2. Product Accumulation Method (累乘法)

Applicable Type

Recurrences of the form:

$$\frac{a_{n+1}}{a_n} = f(n) \quad (a_n \neq 0, f(n) \neq 0)$$

where the product $\prod_{k=1}^{n-1} f(k)$ simplifies via cancellation or known multiplicative formulas.

Algorithm

Write out successive ratios for $k=1, 2, \dots, n-1$:

$$\frac{a_2}{a_1} = f(1), \quad \frac{a_3}{a_2} = f(2), \quad \dots, \quad \frac{a_n}{a_{n-1}} = f(n-1)$$

Multiplying all $n-1$ ratios together yields:

$$\frac{a_n}{a_1} = \prod_{k=1}^{n-1} f(k) \implies a_n = a_1 \cdot \prod_{k=1}^{n-1} f(k) \quad (n \ge 2)$$

3. First-Order Linear Recurrence $a_{n+1} = p a_n + q$ (待定系数构造法)

When $p = 1$, the relation reduces to $a_{n+1} - a_n = q$ (arithmetic sequence). When $q = 0$, it reduces to $a_{n+1} = p a_n$ (geometric sequence). When $p \neq 1$ and $q \neq 0$, we construct an auxiliary geometric sequence.

Undetermined Coefficient Transformation

Assume the recurrence can be written in the form:

$$a_{n+1} - \lambda = p(a_n - \lambda)$$

Expanding this equation: $a_{n+1} = p a_n + \lambda(1 - p)$. Comparing coefficients with $a_{n+1} = p a_n + q$ yields:

$$\lambda(1 - p) = q \implies \lambda = \frac{q}{1 - p}$$

Define the auxiliary sequence $b_n = a_n - \lambda$. Then $b_1 = a_1 - \lambda$, and $b_{n+1} = p b_n$. Thus, $(b_n)$ is a geometric sequence with initial term $b_1$ and common ratio $p$:

$$b_n = b_1 \cdot p^{n-1} = (a_1 - \lambda) p^{n-1}$$

Substituting back $a_n = b_n + \lambda$ gives the general term:

$$a_n = (a_1 - \lambda) p^{n-1} + \lambda = \left(a_1 - \frac{q}{1-p}\right) p^{n-1} + \frac{q}{1-p}$$

4. Higher Recurrence Structures: $a_{n+1} = p a_n + f(n)$

5. Sequence Determination via $S_n$ and $a_n$ Relations

In many Gaokao problems, the defining condition involves both $S_n$ and $a_n$.

Fundamental Bridge

$$a_n = \begin{cases} S_1, \& n = 1 \\ S_n - S_{n-1}, \& n \ge 2 \end{cases}$$

Core Resolution Pathways:

1. Eliminate $S_n$ to find $a_n$ relation: Write the relation for index $n$: $f(S_n, a_n) = 0$. Write the relation for index $n-1$ ($n \ge 2$): $f(S_{n-1}, a_{n-1}) = 0$. Subtract the two equations and substitute $S_n - S_{n-1} = a_n$ to obtain a pure recurrence relation in $a_n$.
2. Eliminate $a_n$ to find $S_n$ relation: Substitute $a_n = S_n - S_{n-1}$ ($n \ge 2$) directly into the given equation to obtain a recurrence relation in $S_n$, find $S_n$, and then recover $a_n = S_n - S_{n-1}$.

Worked Gaokao Exam Examples

Example 1 (Undetermined Coefficients for $a_{n+1} = p a_n + q$)

Problem: A sequence $(a_n)$ satisfies $a_1 = 2$ and $a_{n+1} = 3 a_n - 2$ for all $n \ge 1$.

1. Show that $(a_n - 1)$ is a geometric sequence and state its common ratio.
2. Find the general term formula $a_n$.

Solution:

1. We construct $\lambda$ such that $a_{n+1} - \lambda = 3(a_n - \lambda)$. Expanding gives $a_{n+1} = 3a_n - 2\lambda$. Comparing with $a_{n+1} = 3a_n - 2$:

   $$-2\lambda = -2 \implies \lambda = 1$$

   Thus, $a_{n+1} - 1 = 3(a_n - 1)$ for all $n \ge 1$. Calculate the initial term of $(a_n - 1)$:

   $$b_1 = a_1 - 1 = 2 - 1 = 1$$

   Since $\frac{a_{n+1} - 1}{a_n - 1} = 3$ is a constant ratio, $(a_n - 1)$ is a geometric sequence with initial term $1$ and common ratio $q = 3$.

2. The general term of the auxiliary sequence is:

   $$a_n - 1 = 1 \cdot 3^{n-1} = 3^{n-1}$$

   Adding 1 gives the general term $a_n$:

   $$a_n = 3^{n-1} + 1$$

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Example 2 (General Term from $S_n$ Recurrence Relation)

Problem: Let $(a_n)$ be a sequence with partial sum $S_n$ satisfying $S_n = 2 a_n - 2$ for all $n \ge 1$.

1. Find $a_1$ and $a_2$.
2. Determine the general term formula $a_n$.

Solution:

1. For $n = 1$, $S_1 = a_1$, so:

   $$a_1 = 2a_1 - 2 \implies a_1 = 2$$

   For $n = 2$, $S_2 = a_1 + a_2 = 2 + a_2$. Using $S_2 = 2a_2 - 2$:

   $$2 + a_2 = 2a_2 - 2 \implies a_2 = 4$$

2. For $n \ge 2$, we have:

   $$S_n = 2a_n - 2$$ $$S_{n-1} = 2a_{n-1} - 2$$

   Subtracting the second equation from the first:

   $$S_n - S_{n-1} = 2a_n - 2a_{n-1}$$

   Since $S_n - S_{n-1} = a_n$ for $n \ge 2$:

   $$a_n = 2a_n - 2a_{n-1} \implies a_n = 2a_{n-1} \quad (n \ge 2)$$

   Thus, $(a_n)$ is a geometric sequence with first term $a_1 = 2$ and common ratio $q = 2$. The general term formula is:

   $$a_n = 2 \cdot 2^{n-1} = 2^n$$