Unit 5 · Sequences　数列

By the end of this chapter you can:

1. Find the general term and partial sum of arithmetic and geometric sequences using standard formulas
2. Decode a recurrence relation and compute terms by substitution or pattern recognition
3. Apply 错位相减 (shift-and-subtract) and 裂项相消 (partial-fraction telescoping) to sum non-standard sequences
4. Identify and compute arithmetic and geometric means between two numbers

Exam weight on past CSCA papers: ~10% (4–5 of 48 MCQs).

5.1　Arithmetic Sequences　等差数列

An arithmetic sequence (等差数列) is one where consecutive terms differ by a fixed constant $d$ called the common difference (公差).

$$a_{n+1} - a_n = d \quad \text{for all } n \ge 1$$

Core formulas

🔑 $S_n$ is quadratic in $n$. If a problem gives you $S_n$ and asks for $a_n$, use $a_n = S_n - S_{n-1}$ (for $n \ge 2$) and check $a_1 = S_1$ separately.

Identifying an AP from $S_n$: If $S_n = An^2 + Bn$ (no constant term), then $\{a_n\}$ is arithmetic with $a_1 = A + B$ and $d = 2A$.

⚠️ Trap: If $S_n = An^2 + Bn + C$ with $C \ne 0$, then $\{a_n\}$ is not arithmetic (check with $a_1 = S_1$ vs $a_n = S_n - S_{n-1}$).

Worked Example 5.1.A

In an arithmetic sequence, $a_3 = 7$ and $a_8 = 22$. Find the general term $a_n$ and the sum of the first 20 terms, $S_{20}$.

Solution.

Step 1 — find $d$. Using $a_n = a_1 + (n-1)d$:

$$a_8 - a_3 = (8-1)d - (3-1)d = 5d = 22 - 7 = 15 \implies d = 3.$$

Step 2 — find $a_1$. $a_1 = a_3 - 2d = 7 - 6 = 1$.

Step 3 — write $a_n$.

$$a_n = 1 + (n-1) \cdot 3 = 3n - 2.$$

Step 4 — compute $S_{20}$.

$$S_{20} = \frac{20(a_1 + a_{20})}{2} = \frac{20(1 + 58)}{2} = 10 \times 59 = \boxed{590}.$$

🔑 Shortcut for $a_m - a_k$: the difference of terms $m - k$ steps apart equals $(m-k) \cdot d$. Here $8 - 3 = 5$ steps, so $a_8 - a_3 = 5d$. This saves setting up two separate equations.

5.2　Geometric Sequences　等比数列

A geometric sequence (等比数列) has consecutive terms related by a fixed common ratio (公比) $q \ne 0$:

$$\frac{a_{n+1}}{a_n} = q \quad \text{for all } n \ge 1.$$

Core formulas

⚠️ Never forget to check $q = 1$ separately. Exam questions sometimes embed $q = 1$ as one of the answer choices to catch students who blindly apply the $q \ne 1$ formula.

Equivalent sum formula: $S_n = \dfrac{a_1 - a_n q}{1 - q}$ (for $q \ne 1$) — useful when you know the last term but not $n$.

Worked Example 5.2.A

A geometric sequence has $a_1 = 3$ and $a_4 = 24$. Find $q$, the general term, and $S_5$.

Solution.

Step 1 — find $q$. $a_4 = a_1 q^{3} = 3q^3 = 24 \implies q^3 = 8 \implies q = 2.$

Step 2 — write $a_n$.

$$a_n = 3 \cdot 2^{n-1}.$$

Step 3 — compute $S_5$.

$$S_5 = \frac{3(1 - 2^5)}{1 - 2} = \frac{3(1 - 32)}{-1} = \frac{3(-31)}{-1} = \boxed{93}.$$

🔑 Symmetric-product property: $a_m \cdot a_n = a_p \cdot a_q$ whenever $m + n = p + q$. In particular, $a_k^2 = a_{k-j} \cdot a_{k+j}$ (the middle term squared equals the product of symmetric neighbours). This mirrors the AP property $2a_k = a_{k-j} + a_{k+j}$.

⚠️ Sign of $q$ matters for alternating sequences. If $q < 0$, terms alternate in sign. CSCA questions may ask which term is first negative or first exceeds a value — track sign carefully.

5.3　Recursive Sequences　递推数列

A recurrence relation (递推关系) defines each term from one or more previous terms. CSCA typically tests two types.

Type 1 — Linear first-order: $a_{n+1} = p\,a_n + c$

If $c = 0$: the sequence is geometric with ratio $p$.

If $c \ne 0$: introduce the fixed point $L$ satisfying $L = pL + c$, giving $L = \dfrac{c}{1-p}$. Then $b_n = a_n - L$ is geometric with ratio $p$.

$$a_n = p^{n-1}(a_1 - L) + L \qquad \text{where } L = \frac{c}{1-p}.$$

Type 2 — Read-pattern recurrences

Some recurrences are best solved by computing the first few terms and spotting a pattern (periodicity, AP, GP).

Worked Example 5.3.A

Given $a_1 = 1$ and $a_{n+1} = 2a_n + 3$, find a closed-form expression for $a_n$.

Solution.

Find fixed point: $L = pL + c \Rightarrow L = 2L + 3 \Rightarrow L = -3$.

Set $b_n = a_n - (-3) = a_n + 3$. Then:

$$b_{n+1} = a_{n+1} + 3 = (2a_n + 3) + 3 = 2(a_n + 3) = 2b_n.$$

So $\{b_n\}$ is geometric with ratio $2$ and $b_1 = a_1 + 3 = 4$.

$$b_n = 4 \cdot 2^{n-1} = 2^{n+1} \implies a_n = b_n - 3 = 2^{n+1} - 3.$$

$$\boxed{a_n = 2^{n+1} - 3}$$

Check: $a_1 = 4 - 3 = 1$ ✓; $a_2 = 2(1) + 3 = 5 = 8 - 3$ ✓.

⚠️ Trap: When $p = 1$, the recurrence $a_{n+1} = a_n + c$ is arithmetic (common difference $c$), not geometric. The fixed-point method requires $p \ne 1$.

🔑 Pattern-spotting: if a recurrence looks like $a_{n+2} = a_{n+1} + a_n$ (Fibonacci-type), just list the first several terms. CSCA problems of this type always ask for $a_5$ or $a_6$, so 6 terms suffice.

5.4　Sum Techniques　数列求和方法

Standard AP and GP formulas don't cover products like $n \cdot 2^n$ or terms like $\dfrac{1}{n(n+1)}$. Two classic methods handle these.

Method 1 — 错位相减法 (Shift-and-Subtract)

Used for sums of the form $\displaystyle\sum (\text{polynomial in } n) \cdot q^n$ where $q \ne 1$.

Idea: write $S_n$, then multiply by $q$ to get $qS_n$, subtract, and most terms cancel when you shift the second series by one index.

Worked Example 5.4.A — 错位相减

Find $S_n = \displaystyle\sum_{k=1}^{n} k \cdot 2^k = 1\cdot2 + 2\cdot4 + 3\cdot8 + \cdots + n\cdot 2^n$.

Solution.

Write the two shifted rows:

$$S_n = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \cdots + (n-1)\cdot 2^{n-1} + n\cdot 2^n \tag{1}$$

$$2S_n = \phantom{1\cdot2^1 +{}} 1\cdot2^2 + 2\cdot2^3 + \cdots + (n-2)\cdot 2^{n-1} + (n-1)\cdot 2^n + n\cdot 2^{n+1} \tag{2}$$

Subtract $(1) - (2)$:

$$-S_n = 1\cdot2^1 + (2-1)\cdot2^2 + (3-2)\cdot2^3 + \cdots + \bigl(n-(n-1)\bigr)\cdot 2^n - n\cdot 2^{n+1}$$

$$-S_n = (2 + 2^2 + 2^3 + \cdots + 2^n) - n\cdot 2^{n+1}$$

The parenthesised sum is a GP with first term 2, ratio 2, $n$ terms:

$$2 + 2^2 + \cdots + 2^n = \frac{2(2^n - 1)}{2 - 1} = 2^{n+1} - 2.$$

Therefore:

$$-S_n = (2^{n+1} - 2) - n\cdot 2^{n+1} = (1-n)\cdot 2^{n+1} - 2.$$

$$\boxed{S_n = (n-1)\cdot 2^{n+1} + 2}$$

Check $n=1$: $(0)\cdot4 + 2 = 2 = 1\cdot2^1$ ✓. Check $n=2$: $(1)\cdot8 + 2 = 10 = 2 + 8$ ✓.

🔑 错位相减 recipe in 3 steps:

1. Write $S_n$, then write $q \cdot S_n$ shifted right by one term.
2. Subtract. The "coefficient differences" in the middle all equal 1, leaving a pure GP.
3. Sum the GP, then solve for $S_n$.

Method 2 — 裂项相消法 (Partial-Fraction Telescoping)

Used when each general term can be split into a difference of two simpler fractions whose successive pieces cancel.

Key identities to recognise:

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, \qquad \frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right).$$

Worked Example 5.4.B — 裂项相消

Find $S_n = \displaystyle\sum_{k=1}^{n} \dfrac{1}{k(k+1)}$.

Solution.

Split each term:

$$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$$

Write out the telescoping sum:

$$S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right).$$

All interior fractions cancel in adjacent pairs. Only the very first and very last survive:

$$S_n = 1 - \frac{1}{n+1} = \boxed{\dfrac{n}{n+1}}.$$

🔑 Telescoping diagnostic: if the general term is a rational function of $n$ whose denominator factors into two linear terms differing by a constant $k$, apply the $\dfrac{1}{k}\!\left(\dfrac{1}{n}-\dfrac{1}{n+k}\right)$ identity immediately.

⚠️ Three-factor denominators: $\dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2}\!\left(\dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\right)$. One more layer of telescoping, same idea.

5.5　Arithmetic and Geometric Means　等差中项与等比中项

Arithmetic mean　等差中项

If $a$, $M$, $b$ are three consecutive terms of an AP, then $M$ is the arithmetic mean (等差中项) of $a$ and $b$:

$$2M = a + b \implies M = \frac{a+b}{2}.$$

Equivalently, $M - a = b - M$ (equal common difference on both sides).

Geometric mean　等比中项

If $a$, $G$, $b$ are three consecutive terms of a GP, then $G$ is the geometric mean (等比中项) of $a$ and $b$:

$$G^2 = ab \implies G = \pm\sqrt{ab}.$$

⚠️ Key constraint: $ab > 0$ is required for real $G$ (the two numbers must have the same sign). When the problem specifies $G > 0$, take only the positive square root.

⚠️ Do not confuse the two means. A common CSCA trap: the exam gives $a + b$ and $ab$ and asks which of $a$, $b$ is the arithmetic vs geometric mean of two other numbers. Write both conditions and test each option.

AM–GM inequality (bonus fact for "compare" questions)

For positive numbers: $\dfrac{a + b}{2} \ge \sqrt{ab}$, with equality iff $a = b$.

Worked Example 5.5.A

Three positive numbers $a$, $G$, $b$ satisfy both the AP condition and the GP condition simultaneously. What can you conclude?

Solution.

AP condition: $2G = a + b$.

GP condition: $G^2 = ab$.

Treating $a$ and $b$ as unknowns with known sum $s = 2G$ and product $p = G^2$, they are roots of:

$$t^2 - 2G\,t + G^2 = (t - G)^2 = 0.$$

The only solution is $a = b = G$.

$$\boxed{a = b = G \text{ — all three numbers are equal.}}$$

🔑 Shortcut rule: a sequence that is simultaneously arithmetic and geometric must be constant.

Try it!　自测练习

Q1. An AP has $a_5 = 11$ and $a_{12} = 32$. Find $a_1$ and $d$.

Q2. A GP has $a_2 = 6$ and $a_5 = 48$. Find $q$ and $S_6$.

Q3. Given $a_1 = 2$ and $a_{n+1} = 3a_n - 4$, find a closed-form expression for $a_n$.

Q4. Use 裂项相消 to evaluate $\displaystyle\sum_{k=1}^{10} \frac{1}{k(k+2)}$.

Q5. Two positive numbers have arithmetic mean $5$ and geometric mean $4$. Find the two numbers.

📌 Chapter summary

Topic	Key formulas / methods
Arithmetic sequences 等差数列	$a_n = a_1 + (n-1)d$; $S_n = \tfrac{n(a_1+a_n)}{2}$; $S_n$ is quadratic in $n$
Geometric sequences 等比数列	$a_n = a_1 q^{n-1}$; $S_n = \tfrac{a_1(1-q^n)}{1-q}$ ($q\ne1$); check $q=1$ separately
Recursive sequences 递推数列	$a_{n+1} = pa_n + c$: find fixed point $L = c/(1-p)$, reduce to GP; $p = 1$ gives AP
错位相减 (shift-subtract)	$S_n$ with $(\text{poly})\cdot q^n$: write $S_n$ and $qS_n$, subtract, sum the remaining GP
裂项相消 (telescoping)	$\tfrac{1}{n(n+k)} = \tfrac{1}{k}\!\left(\tfrac{1}{n}-\tfrac{1}{n+k}\right)$; write all terms, cancel pairs
Arithmetic mean 等差中项	$M = \tfrac{a+b}{2}$; equivalently $2M = a+b$
Geometric mean 等比中项	$G = \pm\sqrt{ab}$ ($ab > 0$); equivalently $G^2 = ab$

What's next → Unit 6 introduces plane vectors, where the dot-product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ echoes the GM identity in a geometric setting.