---
title: "Solve aₙ=2aₙ₋₁+2n²"
subject: "Discrete Structure"
chapter: 3
key: "semester_4/discrete_structure:recurrence-a-n-2a-2n-squared"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/recurrence-a-n-2a-2n-squared
site: BCT Engineering
---

# Solve aₙ=2aₙ₋₁+2n²

Discrete Structure · Chapter 3: Recurrence Relation

- Asked in: 2074 Bhadra, 5 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-a-n-2a-2n-squared
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-3

## Question

Find all solutions of the recurrence relation aₙ = 2aₙ₋₁ + 2n² with initial condition a₁ = 4.

## Short answer

Homogeneous: α·2^n. Particular: -2n^2 - 8n - 12. General solution: a_n = α·2^n - 2n^2 - 8n - 12. With a1=4, α=13, so a_n = 13·2^n - 2n^2 - 8n - 12.

## Answer

The relation aₙ = 2aₙ₋₁ + 2n² is linear, non-homogeneous, of order 1.

#### Step 1. Homogeneous part
aₙ = 2aₙ₋₁ gives r = 2, so aₙ⁽ʰ⁾ = α·2ⁿ.

#### Step 2. Particular solution
F(n) = 2n² is a polynomial of degree 2 and 1 is not a root, so try aₙ⁽ᵖ⁾ = pn² + qn + r.
```text
pn² + qn + r = 2[p(n-1)² + q(n-1) + r] + 2n²
             = 2pn² - 4pn + 2p + 2qn - 2q + 2r + 2n²
n²:     p = 2p + 2            ->  p = -2
n:      q = -4p + 2q          ->  q = 4p = -8
const:  r = 2p - 2q + 2r      ->  r = -2p + 2q = 4 - 16 = -12
```
aₙ⁽ᵖ⁾ = −2n² − 8n − 12.

#### Step 3. General solution (all solutions)
aₙ = α·2ⁿ − 2n² − 8n − 12, where α is any constant.

#### Step 4. Initial condition a₁ = 4
2α − 2 − 8 − 12 = 4, so 2α = 26 and α = 13.

**Answer: aₙ = 13·2ⁿ − 2n² − 8n − 12**

Check: a₁ = 26 − 2 − 8 − 12 = 4; a₂ = 52 − 8 − 16 − 12 = 16 = 2(4) + 2(4) ✓; a₃ = 104 − 18 − 24 − 12 = 50 = 2(16) + 18 ✓.

## Check yourself

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

1. What is the particular solution guess for F(n) = 2n^2?
   A. pn^2 + qn + r
   B. c·2^n
   C. pn^2 only
   D. c·n·2^n
   Answer: A. A degree-2 polynomial, since 1 is not a root. (Key point: Trial form)

2. Comparing n^2 coefficients, what is p?
   A. p = 2
   B. p = −2
   C. p = 1
   D. p = −1
   Answer: B. p = 2p + 2 gives p = −2. (Key point: Coefficient p)

3. Which is the general solution?
   A. a_n = α·2^n + 2n^2
   B. a_n = α·2^n − 2n^2 + 8n + 12
   C. a_n = α·2^n − 2n^2 − 8n − 12
   D. a_n = −2n^2 − 8n − 12
   Answer: C. Homogeneous α·2^n plus the particular solution. (Key point: General solution)

4. With a1 = 4, what is α?
   A. 26
   B. 4
   C. −13
   D. 13
   Answer: D. 2α − 2 − 8 − 12 = 4 gives α = 13. (Key point: Value of α)

## Close variants set in other exams

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