7.2 Geometric Sequences and Sums

7.2 Geometric Sequences and Sums

1. Mathematical Definition and Recursive Structure

A geometric sequence (等比数列) is a sequence of non-zero real numbers $(a_n)$ where the ratio of any term to its immediate predecessor is a constant non-zero real number. Formally, a sequence $(a_n)$ with $a_1 \neq 0$ is a geometric sequence if and only if there exists $q \in \mathbb{R} \setminus \{0\}$, called the common ratio (公比), such that:

$$\frac{a_{n+1}}{a_n} = q, \quad \forall n \in \mathbb{N}^*$$

Equivalently, the sequence satisfies the non-zero three-term relation:

$$a_n^2 = a_{n-1} \cdot a_{n+1}, \quad \forall n \ge 2, n \in \mathbb{N}^*$$

General Term Derivation

To derive the general term $a_n$ from $a_1$ and $q$, we employ the product accumulation method (累乘法). Writing successive ratios:

$$\begin{aligned} \frac{a_2}{a_1} \&= q \\ \frac{a_3}{a_2} \&= q \\ \&\vdots \\ \frac{a_n}{a_{n-1}} \&= q \end{aligned}$$

Multiplying all $n-1$ equations together telescopically cancels intermediate terms:

$$\frac{a_2}{a_1} \cdot \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = \prod_{k=1}^{n-1} q \implies \frac{a_n}{a_1} = q^{n-1} \implies a_n = a_1 q^{n-1}$$

2. Discrete Exponential Function Interpretation

Writing $a_n = \frac{a_1}{q} \cdot q^n$ demonstrates that a geometric sequence corresponds to a discrete exponential function $f(n) = C \cdot q^n$ defined on $\mathbb{N}^*$.

3. Core Term Index Properties

Theorem (Multiplicative Index Equivalence)

If $(a_n)$ is a geometric sequence and $m, n, p, q \in \mathbb{N}^*$ satisfy $m + n = p + q$, then:

$$a_m \cdot a_n = a_p \cdot a_q$$

Proof: Expressing each term using the general term formula $a_k = a_1 q^{k-1}$:

$$a_m \cdot a_n = (a_1 q^{m-1})(a_1 q^{n-1}) = a_1^2 q^{m+n-2}$$ $$a_p \cdot a_q = (a_1 q^{p-1})(a_1 q^{q-1}) = a_1^2 q^{p+q-2}$$

Since $m + n = p + q$, the powers of $q$ are equal, proving identity. $\blacksquare$

Corollary: Geometric Mean Property

For any term index $k \ge 2$, setting $m = k-1$ and $n = k+1$ yields $(k-1) + (k+1) = 2k$, so:

$$a_{k-1} \cdot a_{k+1} = a_k^2 \implies |a_k| = \sqrt{a_{k-1} a_{k+1}}$$

Gaokao Warning: If $a, G, b$ form a geometric sequence, then $G^2 = ab$. Thus $G = \pm \sqrt{ab}$. The geometric mean can be positive or negative depending on the common ratio $q$. Assuming $G$ must be positive is one of the most frequent traps in Gaokao objective questions!

4. Partial Sum Formulas ($S_n$)

The partial sum $S_n = a_1 + a_2 + \dots + a_n$ of a geometric sequence requires careful case distinction depending on whether $q = 1$ or $q \neq 1$.

Derivation via Multiplication-Subtraction Method

When $q \neq 1$, write $S_n$ and multiply by $q$:

$$\begin{aligned} S_n \&= a_1 + a_1 q + a_1 q^2 + \dots + a_1 q^{n-1} \\ q S_n \&= a_1 q + a_1 q^2 + a_1 q^3 + \dots + a_1 q^n \end{aligned}$$

Subtracting the second equation from the first cancels all middle terms:

$$(1 - q) S_n = a_1 - a_1 q^n = a_1 (1 - q^n)$$

Since $q \neq 1$, we divide by $1 - q$:

$$S_n = \frac{a_1 (1 - q^n)}{1 - q} = \frac{a_1 - a_n q}{1 - q} \quad (q \neq 1)$$

If $q = 1$, all terms equal $a_1$, so:

$$S_n = n a_1 \quad (q = 1)$$

Structural Functional Form of Geometric $S_n$

Rewriting $S_n$ for $q \neq 1$:

$$S_n = \frac{a_1}{1-q} - \frac{a_1}{1-q} q^n = A - A q^n = A q^n + B, \quad \text{where } A = -\frac{a_1}{1-q}, B = \frac{a_1}{1-q}$$

Key Gaokao Insight: A sequence $(a_n)$ has partial sum $S_n = A q^n + B$ representing a geometric sequence if and only if $A + B = 0$ (i.e., $B = -A$) and $q \neq 0, q \neq 1$. If $A + B \neq 0$, then $a_1 = S_1 = A q + B$, whereas for $n \ge 2$, $a_n = A(q-1)q^{n-1}$, meaning the geometric pattern begins only at $n=2$.

5. Partial Block Sum Property for Geometric Sequences

For a geometric sequence $(a_n)$ with sum $S_n$, consider consecutive blocks of length $k$:

$$T_1 = S_k = a_1 + a_2 + \dots + a_k$$ $$T_2 = S_{2k} - S_k = a_{k+1} + a_{k+2} + \dots + a_{2k} = q^k (a_1 + a_2 + \dots + a_k) = q^k T_1$$ $$T_3 = S_{3k} - S_{2k} = a_{2k+1} + \dots + a_{3k} = q^k T_2 = q^{2k} T_1$$

Provided $T_1 \neq 0$ and $q^k \neq -1$, the block sums $(T_1, T_2, T_3, \dots)$ form a geometric sequence with common ratio $Q = q^k$.

6. Infinite Geometric Series Convergence

For an infinite geometric series $\sum_{k=1}^\infty a_1 q^{k-1}$, the limit of partial sums as $n \to \infty$ is:

$$S_\infty = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a_1(1-q^n)}{1-q}$$

• If $|q| < 1$, then $\lim_{n \to \infty} q^n = 0$, so the series converges to $S_\infty = \frac{a_1}{1-q}$.
• If $|q| \ge 1$, $\lim_{n \to \infty} q^n$ does not exist or is infinite, so the series diverges.

Worked Gaokao Exam Examples

Example 1 (Term Determination and Geometric Properties)

Problem: In a geometric sequence $(a_n)$ of positive real numbers, $a_2 a_4 = 16$ and $a_1 + a_3 = 10$.

1. Find the general term formula for $a_n$.
2. Calculate the sum $S_5$.

Solution:

1. Since all terms are positive, $a_1 > 0$ and $q > 0$. Using the term index property: $a_2 a_4 = a_3^2 = 16$. Because terms are positive, $a_3 = \sqrt{16} = 4$. We are given $a_1 + a_3 = 10$, so $a_1 + 4 = 10 \implies a_1 = 6$. Since $a_3 = a_1 q^2$, we have $6 q^2 = 4 \implies q^2 = \frac{2}{3}$. Since $q > 0$, $q = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$. The general term formula is:

   $$a_n = 6 \cdot \left(\frac{\sqrt{6}}{3}\right)^{n-1}$$

2. To find $S_5$: Computing term by term:

   $$a_1 = 6, \quad a_2 = 6 \sqrt{\frac{2}{3}} = 2\sqrt{6}, \quad a_3 = 4, \quad a_4 = 4 \sqrt{\frac{2}{3}} = \frac{4\sqrt{6}}{3}, \quad a_5 = \frac{8}{3}$$

   Summing terms directly:

   $$S_5 = 6 + 4 + \frac{8}{3} + \left(2\sqrt{6} + \frac{4\sqrt{6}}{3}\right) = \frac{38}{3} + \frac{10\sqrt{6}}{3} = \frac{38 + 10\sqrt{6}}{3}$$

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Example 2 (Block Sum Property of Geometric Sequences)

Problem: Let $(a_n)$ be a geometric sequence with common ratio $q \neq -1$. If $S_3 = 7$ and $S_6 = 63$, find:

1. The common ratio $q$.
2. The general term formula $a_n$.

Solution:

1. Let $T_1 = S_3 = 7$, and $T_2 = S_6 - S_3 = 63 - 7 = 56$. By the block partial sum property, $T_1$ and $T_2$ are consecutive blocks of size $k=3$, so:

   $$\frac{T_2}{T_1} = q^3 \implies q^3 = \frac{56}{7} = 8$$

   Since $q \in \mathbb{R}$, $q = \sqrt[3]{8} = 2$.

2. Using $S_3 = \frac{a_1(1-q^3)}{1-q} = 7$:

   $$\frac{a_1(1 - 8)}{1 - 2} = \frac{-7 a_1}{-1} = 7 a_1 = 7 \implies a_1 = 1$$

   Thus, the general term is:

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