7.3 Recurrence Relations and General Term Determination
Key Takeaways
- For additive recurrences a_{n+1} - a_n = f(n), the difference accumulation method gives a_n = a_1 + Sum_{k=1}^{n-1} f(k).
- For multiplicative recurrences a_{n+1}/a_n = f(n), the product accumulation method gives a_n = a_1 * Prod_{k=1}^{n-1} f(k).
- First-order linear recurrences a_{n+1} = p a_n + q (p != 1) are solved by undetermined coefficient transformation a_{n+1} - lambda = p(a_n - lambda) with lambda = q / (1-p).
- When given S_n relations, apply a_n = S_n - S_{n-1} for n >= 2, always verifying whether the initial term a_1 = S_1 conforms to the general pattern.
- Recurrences involving variable additions like a_{n+1} = p a_n + k r^n can be linearized by dividing by r^{n+1} to create an auxiliary sequence.
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:
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 $n-1$:
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:
Expanding this equation: $a_{n+1} = p a_n + \lambda(1 - p)$. Comparing coefficients with $a_{n+1} = p a_n + q$ yields:
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 $p$:
Substituting back $a_n = b_n + \lambda$ gives the general term:
4. Higher Recurrence Structures: $a_{n+1} = p a_n + f(n)$
| Function Type $f(n)$ | Auxiliary Transformation Strategy | Resulting Auxiliary Sequence |
|---|---|---|
| $f(n) = k r^n$ ($r \neq p$) | Divide both sides by $r^{n+1}$: $\frac{a_{n+1}}{r^{n+1}} = \frac{p}{r} \frac{a_n}{r^n} + \frac{k}{r}$ | Let $b_n = \frac{a_n}{r^n}$, reducing to $b_{n+1} = p' b_n + q'$ |
| $f(n) = k p^n$ ($r = p$) | Divide both sides by $p^{n+1}$: $\frac{a_{n+1}}{p^{n+1}} = \frac{a_n}{p^n} + \frac{k}{p}$ | Let $b_n = \frac{a_n}{p^n}$, giving AP $b_{n+1} - b_n = \frac{k}{p}$ |
| $f(n) = A n + B$ | Construct $a_{n+1} + X(n+1) + Y = p(a_n + Xn + Y)$ | Match coefficients to solve for $X$ and $Y$ |
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
Core Resolution Pathways:
- 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 $a_n$.
- 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 $n \ge 1$.
- Show that $(a_n - 1)$ is a geometric sequence and state its common ratio.
- Find the general term formula $a_n$.
Solution:
-
We construct $\lambda$ such that $a_{n+1} - \lambda = 3(a_n - \lambda)$. Expanding gives $a_{n+1} = 3a_n - 2\lambda$. Comparing with $a_{n+1} = 3a_n - 2$: Thus, $a_{n+1} - 1 = 3(a_n - 1)$ for all $n \ge 1$. Calculate the initial term of $(a_n - 1)$: 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 $q = 3$.
-
The general term of the auxiliary sequence is: Adding 1 gives the general term $a_n$:
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 $n \ge 1$.
- Find $a_1$ and $a_2$.
- Determine the general term formula $a_n$.
Solution:
-
For $n = 1$, $S_1 = a_1$, so: For $n = 2$, $S_2 = a_1 + a_2 = 2 + a_2$. Using $S_2 = 2a_2 - 2$:
-
For $n \ge 2$, we have: Subtracting the second equation from the first: Since $S_n - S_{n-1} = a_n$ for $n \ge 2$: Thus, $(a_n)$ is a geometric sequence with first term $a_1 = 2$ and common ratio $q = 2$. The general term formula is:
Given a_1 = 2 and a_{n+1} = 3a_n - 2, what is the general term a_n?
A sequence satisfies a_1 = 1 and a_{n+1} - a_n = 2n. What is the value of a_10?
For a sequence (a_n) with partial sum S_n, if S_n = 3^n - 1, what is a_n for n >= 1?