---
title: "Solve aₙ=5aₙ₋₁−6aₙ₋₂+2ⁿ"
subject: "Discrete Structure"
chapter: 3
key: "semester_4/discrete_structure:recurrence-5a-6a-2n-a0-1-a1-4"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/recurrence-5a-6a-2n-a0-1-a1-4
site: BCT Engineering
---

# Solve aₙ=5aₙ₋₁−6aₙ₋₂+2ⁿ

Discrete Structure · Chapter 3: Recurrence Relation

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/recurrence-5a-6a-2n-a0-1-a1-4
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-3

## Question

Find all the solutions of the recurrence relation: aₙ = 5aₙ₋₁ − 6aₙ₋₂ + 2ⁿ with initial conditions a₀ = 1 and a₁ = 4.

## Short answer

Homogeneous solution α·2^n + β·3^n; since 2 is a root of the characteristic equation, use particular solution -n·2^{n+1}. With a_0=1, a_1=4 gives α=-5, β=6. Thus a_n = -5·2^n + 6·3^n - n·2^{n+1}.

## Answer

aₙ = 5aₙ₋₁ − 6aₙ₋₂ + 2ⁿ is linear, non-homogeneous, of order 2.

#### Step 1. Homogeneous part
r² − 5r + 6 = 0 gives r = 2, 3, so aₙ⁽ʰ⁾ = α·2ⁿ + β·3ⁿ.

#### Step 2. Particular solution
F(n) = 2ⁿ, and 2 is a (simple) root of the characteristic equation, so the usual guess c·2ⁿ fails; try aₙ⁽ᵖ⁾ = c·n·2ⁿ.
```text
c n 2ⁿ = 5c(n-1)2ⁿ⁻¹ - 6c(n-2)2ⁿ⁻² + 2ⁿ          (divide by 2ⁿ⁻²)
4cn    = 10c(n-1) - 6c(n-2) + 4
4cn    = 4cn + 2c + 4
c = -2
```
aₙ⁽ᵖ⁾ = −2n·2ⁿ = −n·2ⁿ⁺¹.

#### Step 3. All solutions
aₙ = α·2ⁿ + β·3ⁿ − n·2ⁿ⁺¹

#### Step 4. Initial conditions
```text
a₀ = 1:  α + β = 1
a₁ = 4:  2α + 3β - 4 = 4  ->  2α + 3β = 8
```
From the first, α = 1 − β; then 2 − 2β + 3β = 8 so β = 6 and α = −5.

**Answer: aₙ = −5·2ⁿ + 6·3ⁿ − n·2ⁿ⁺¹**

Check: a₂ = 5(4) − 6(1) + 4 = 18, formula: −20 + 54 − 16 = 18 ✓. a₃ = 5(18) − 24 + 8 = 74, formula: −40 + 162 − 48 = 74 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the roots of r^2 − 5r + 6 = 0?
   A. −2 and −3
   B. 2 and 3
   C. 1 and 6
   D. 2 and 6
   Answer: B. (r−2)(r−3) = 0. (Key point: Roots)

2. Why use c·n·2^n for F(n) = 2^n?
   A. 2 is a double root
   B. 3 is a root
   C. 2 is a simple root of the characteristic equation
   D. F(n) is linear
   Answer: C. Multiply by n when the guess duplicates a homogeneous term. (Key point: Trial form)

3. What is c?
   A. 2
   B. −1
   C. 1
   D. −2
   Answer: D. 4cn = 4cn + 2c + 4 gives c = −2. (Key point: Value of c)

4. Using a0 = 1, a1 = 4, what are α and β?
   A. α = −5, β = 6
   B. α = 6, β = −5
   C. α = 1, β = 0
   D. α = −1, β = 2
   Answer: A. α + β = 1 and 2α + 3β = 8. (Key point: Constants)

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