Study Notes – Reasoning Ability – Series
Audience: Chandigarh Patwari / Lekhpal Examination
Level: MIT‑style University (Rigorous, First‑Principles, Edge Cases)**
Why series?
In reasoning tests the series section evaluates your ability to spot hidden patterns, extract governing rules, and extrapolate future terms. A deep understanding of the mathematics behind sequences and series gives you a systematic toolkit that eliminates guesswork.
1. Foundations
| Symbol | Meaning | Example |
|---|---|---|
| (a_n) | (n)-th term of a sequence | (a_1=2, a_2=5, a_3=8,\dots) |
| (S_n) | Sum of first (n) terms of a series | (S_4 = a_1+a_2+a_3+a_4) |
| (r) | Common ratio (geometric) | (r=\frac{a_{n+1}}{a_n}) |
| (d) | Common difference (arithmetic) | (d=a_{n+1}-a_n) |
| (\Delta a_n = a_{n+1}-a_n) | First difference | |
| (\Delta^2 a_n = \Delta(\Delta a_n)) | Second difference |
First Principle – Every series can be viewed as a list of integers (or rationals) that arise from a simple rule (linear, quadratic, exponential, recursive, etc.). Your task is to discover that rule.
2. Classical Series Types
| Type | Rule | General Term (a_n) | Sum Formula (finite) |
|---|---|---|---|
| Arithmetic | (a_{n+1}=a_n+d) | (a_n=a_1+(n-1)d) | (S_n=\frac{n}{2}(a_1+a_n)) |
| Geometric | (a_{n+1}=a_nr) | (a_n=a_1 r^{,n-1}) | (S_n=a_1\frac{1-r^n}{1-r}) (if (r\neq1)) |
| Harmonic | (a_n=\frac{1}{n}) | (a_n=\frac{1}{n}) | Diverges as (n\to\infty) |
| Quadratic | (\Delta^2 a_n = c) | (a_n=An^2+Bn+C) | (S_n=\frac{n}{6}\bigl[(2A+3B+3C)+(2A-3B+3C)n^2\bigr]) |
| Fibonacci | (a_{n}=a_{n-1}+a_{n-2}) | (a_n=\frac{\phi^n-\psi^n}{\sqrt5}) | (\sum_{k=1}^n a_k = a_{n+2}-1) |
Proof sketch (Arithmetic):
Assume (a_{n+1}=a_n+d). By induction (a_n=a_1+(n-1)d).
Sum: (S_n = a_1 + (a_1+d)+\dots+(a_1+(n-1)d) = n a_1 + d\frac{n(n-1)}2).
Edge Cases
Zero difference/ratio → constant series.
Negative ratio → alternating sign.
Common difference zero → constant series.
3. Advanced Patterns
3.1 Second‑Difference Method (Quadratic Series)
If the first differences (\Delta a_n) are not constant but the second differences (\Delta^2 a_n) are constant, the series is quadratic.
Algorithm
- Compute (\Delta a_n = a_{n+1}-a_n).
- Compute (\Delta^2 a_n = \Delta a_{n+1}-\Delta a_n).
- If (\Delta^2 a_n = c) constant, then (a_n = \frac{c}{2}n^2 + (d-\frac{c}{2})n + a_1).
3.2 Recurrence Relations & Characteristic Equations
Many series satisfy linear homogeneous recurrences: [ a_n = \sum_{k=1}^m p_k a_{n-k}, \quad p_k\in\mathbb{R} ] Solve using the characteristic polynomial: [ \lambda^m - \sum_{k=1}^m p_k \lambda^{m-k}=0 ] Roots (\lambda_i) give general solution: [ a_n = \sum_{i} C_i \lambda_i^n ] Example: Fibonacci → (a_n = a_{n-1}+a_{n-2}) → characteristic (\lambda^2-\lambda-1=0) → (\lambda_{1,2}=\frac{1\pm\sqrt5}{2}).