📈 Sequences & Series

Master arithmetic and geometric sequences, telescoping series, and partial sums. These patterns appear frequently in AMC problems.

🎯 Key Ideas

Sequence: Ordered list of numbers $a_1, a_2, a_3, \ldots$

Series: Sum of sequence terms $S_n = a_1 + a_2 + \cdots + a_n$

Arithmetic: Each term differs by constant amount (common difference $d$)

Geometric: Each term differs by constant ratio (common ratio $r$)

➕ Arithmetic Sequences

Definition

Sequence where $a_{n+1} - a_n = d$ (constant) for all $n$.

General Term

$$a_n = a_1 + (n-1)d$$

Sum of First $n$ Terms

$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}[2a_1 + (n-1)d]$$

Example: Find sum of first 20 terms of arithmetic sequence with $a_1 = 3$ and $d = 4$

Solution:

  1. Find $a_{20}$: $a_{20} = 3 + (20-1) \cdot 4 = 3 + 19 \cdot 4 = 3 + 76 = 79$
  2. Apply sum formula: $S_{20} = \frac{20}{2}(3 + 79) = 10 \cdot 82 = 820$

✖️ Geometric Sequences

Definition

Sequence where $\frac{a_{n+1}}{a_n} = r$ (constant) for all $n$.

General Term

$$a_n = a_1 r^{n-1}$$

Sum of First $n$ Terms

$$S_n = a_1 \frac{1-r^n}{1-r} \text{ (for } r \neq 1\text{)}$$

Infinite Sum (for $|r| < 1$)

$$S_{\infty} = \frac{a_1}{1-r}$$

Example: Find sum of first 10 terms of geometric sequence with $a_1 = 2$ and $r = 3$

Solution:

  • $S_{10} = 2 \cdot \frac{1-3^{10}}{1-3} = 2 \cdot \frac{1-59049}{-2} = 2 \cdot \frac{-59048}{-2} = 2 \cdot 29524 = 59048$

🔄 Telescoping Series

Strategy

Express terms as differences: $a_n = b_n - b_{n+1}$, then: $$\sum_{k=1}^n a_k = b_1 - b_{n+1}$$

Common Pattern

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

Example: Find $\sum_{k=1}^{100} \frac{1}{k(k+1)}$

Solution:

  1. Decompose: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
  2. Apply telescoping:
    • $\sum_{k=1}^{100} \frac{1}{k(k+1)} = \sum_{k=1}^{100} \left(\frac{1}{k} - \frac{1}{k+1}\right)$
    • $= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{100} - \frac{1}{101}\right)$
    • $= 1 - \frac{1}{101} = \frac{100}{101}$

📊 Partial Sums

Arithmetic Partial Sum

$$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$

Geometric Partial Sum

$$S_n = a_1 \frac{1-r^n}{1-r} \text{ (for } r \neq 1\text{)}$$

Example: Find partial sum $S_8$ for sequence $2, 6, 18, 54, \ldots$

Solution:

  1. Identify as geometric: $a_1 = 2$, $r = \frac{6}{2} = 3$
  2. Apply formula: $S_8 = 2 \cdot \frac{1-3^8}{1-3} = 2 \cdot \frac{1-6561}{-2} = 2 \cdot \frac{-6560}{-2} = 6560$

🔢 Special Sequences

Triangular Numbers

$$T_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$

Square Numbers

$$S_n = n^2$$

Fibonacci Sequence

$$F_1 = 1, F_2 = 1, F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 3$$

Example: Find sum of first 10 triangular numbers

Solution:

  • $\sum_{k=1}^{10} T_k = \sum_{k=1}^{10} \frac{k(k+1)}{2} = \frac{1}{2}\sum_{k=1}^{10} (k^2 + k)$
  • $= \frac{1}{2}\left[\sum_{k=1}^{10} k^2 + \sum_{k=1}^{10} k\right]$
  • $= \frac{1}{2}\left[\frac{10 \cdot 11 \cdot 21}{6} + \frac{10 \cdot 11}{2}\right]$
  • $= \frac{1}{2}[385 + 55] = \frac{440}{2} = 220$

🎯 AMC-Style Worked Example

Problem: Find the sum $\sum_{n=1}^{\infty} \frac{2^n + 3^n}{6^n}$.

Solution:

  1. Split the sum: $\sum_{n=1}^{\infty} \frac{2^n + 3^n}{6^n} = \sum_{n=1}^{\infty} \frac{2^n}{6^n} + \sum_{n=1}^{\infty} \frac{3^n}{6^n}$
  2. Simplify each term:
    • $\frac{2^n}{6^n} = \left(\frac{2}{6}\right)^n = \left(\frac{1}{3}\right)^n$
    • $\frac{3^n}{6^n} = \left(\frac{3}{6}\right)^n = \left(\frac{1}{2}\right)^n$
  3. Apply infinite geometric sum formula:
    • $\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$
    • $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$
  4. Add results: $\frac{1}{2} + 1 = \frac{3}{2}$

Answer: $\frac{3}{2}$

🔍 Common Traps & Fixes

Trap: Wrong formula for geometric sum

Fix: Use $S_n = a_1 \frac{1-r^n}{1-r}$ for finite sum, $S_{\infty} = \frac{a_1}{1-r}$ for infinite sum.

Trap: Forgetting to check convergence

Fix: Infinite geometric series converges only when $|r| < 1$.

Trap: Wrong telescoping decomposition

Fix: Check that $a_n = b_n - b_{n+1}$ by expanding the right side.

Trap: Off-by-one errors in indexing

Fix: Be careful with first term ($a_1$ vs $a_0$) and indexing in formulas.

📋 Quick Reference

Arithmetic Sequence

  • General term: $a_n = a_1 + (n-1)d$
  • Sum: $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}[2a_1 + (n-1)d]$

Geometric Sequence

  • General term: $a_n = a_1 r^{n-1}$
  • Finite sum: $S_n = a_1 \frac{1-r^n}{1-r}$ (for $r \neq 1$)
  • Infinite sum: $S_{\infty} = \frac{a_1}{1-r}$ (for $|r| < 1$)

Common Telescoping Patterns

  • $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
  • $\frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$

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