---
title: "Induction: sum of cubes"
subject: "Discrete Structure"
chapter: 1
key: "semester_4/discrete_structure:induction-sum-of-cubes"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/induction-sum-of-cubes
site: BCT Engineering
---

# Induction: sum of cubes

Discrete Structure · Chapter 1: Logic, Induction and Reasoning

- Asked in: 2071 Bhadra, 8 marks
- Video: [Prove by induction, Sum of the first n cubes, 1^3+2^3+3^3+...+n^3](https://www.youtube.com/watch?v=w362XRZy5as) — blackpenredpen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/induction-sum-of-cubes
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-1

## Question

Use mathematical induction to show that 1³ + 2³ + .... + n³ = [n(n+1)/2]² whenever n is a positive integer.

## Short answer

Basis n=1: 1³ = [1·2/2]². Assume sum to k = [k(k+1)/2]². Add (k+1)³, factor to get (k+1)²(k²+4k+4)/4 = [(k+1)(k+2)/2]². Holds by induction.

## Answer

**To prove.** P(n): 1³ + 2³ + … + n³ = [n(n+1)/2]² for all positive integers n.

#### Basis step (n = 1)
- LHS = 1³ = 1.
- RHS = [1·2/2]² = 1.
- P(1) is true.

#### Inductive hypothesis
Assume P(k) is true: 1³ + 2³ + … + k³ = [k(k+1)/2]² = k²(k+1)²/4.

#### Inductive step
Show P(k+1): 1³ + … + k³ + (k+1)³ = [(k+1)(k+2)/2]².

```text
LHS = k²(k+1)²/4 + (k+1)³                 (hypothesis)
    = (k+1)² [ k²/4 + (k+1) ]
    = (k+1)² (k² + 4k + 4)/4
    = (k+1)² (k+2)²/4
    = [(k+1)(k+2)/2]²  = RHS
```

#### Conclusion
P(1) is true and P(k) → P(k+1), so by mathematical induction P(n) is true for all positive integers n. ∎

## Check yourself

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

1. At n = 1, [n(n+1)/2]² equals?
   A. 2
   B. 4
   C. 3
   D. 1
   Answer: D. [1·2/2]² = 1 = 1³. (Key point: Basis)

2. What is added to the hypothesis in the step?
   A. (k+1)²
   B. k³
   C. (k+2)³
   D. (k+1)³
   Answer: D. The (k+1)th cube is (k+1)³. (Key point: Next term)

3. After factoring (k+1)², the bracket is?
   A. k² + (k+1)
   B. k/4 + 1
   C. k²/2 + k
   D. k²/4 + (k+1)
   Answer: D. k²(k+1)²/4 + (k+1)³ = (k+1)²[k²/4 + (k+1)]. (Key point: Factoring)

4. k² + 4k + 4 equals?
   A. (k+4)²
   B. (k+1)²
   C. k² + 4
   D. (k+2)²
   Answer: D. It is a perfect square, giving (k+1)²(k+2)²/4. (Key point: Perfect square)

## Close variants set in other exams

- [Induction: sum of squares](https://www.bctengineer.com/question.md?key=semester_4%2Fdiscrete_structure%3Ainduction-sum-of-squares) — Discrete Structure, chapter 1, asked 2×

## Related in Discrete Structure

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