---
title: "Induction: sum of n(n+1)(n+2)"
subject: "Discrete Structure"
chapter: 1
key: "semester_4/discrete_structure:induction-product-n-n1-n2"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/induction-product-n-n1-n2
site: BCT Engineering
---

# Induction: sum of n(n+1)(n+2)

Discrete Structure · Chapter 1: Logic, Induction and Reasoning

- Asked in: 2070 Bhadra, 8 marks
- Video: [Proof by Mathematical Induction - How to do a Mathematical Induction Proof ( Example 1 )](https://www.youtube.com/watch?v=dMn5w4_ztSw) — Learn Math Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/induction-product-n-n1-n2
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-1

## Question

Prove by Mathematical Induction: 1.2.3 + 2.3.4 + 3.4.5 + ........ + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4

## Short answer

Basis n=1: 1·2·3 = 1·2·3·4/4. Assume sum to k = k(k+1)(k+2)(k+3)/4. Add (k+1)(k+2)(k+3), factor (k+1)(k+2)(k+3) to get (k+1)(k+2)(k+3)(k+4)/4. True for all n≥1.

## Answer

**To prove.** P(n): 1·2·3 + 2·3·4 + … + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4 for all n ≥ 1.

#### Basis step (n = 1)
- LHS = 1·2·3 = 6.
- RHS = 1·2·3·4/4 = 6.
- P(1) is true.

#### Inductive hypothesis
Assume P(k): 1·2·3 + … + k(k+1)(k+2) = k(k+1)(k+2)(k+3)/4.

#### Inductive step
Show P(k+1): the sum up to (k+1)(k+2)(k+3) equals (k+1)(k+2)(k+3)(k+4)/4.

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

#### Conclusion
By mathematical induction, P(n) is true for every positive integer n. ∎

## Check yourself

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

1. At n = 1 both sides equal?
   A. 24
   B. 1
   C. 3
   D. 6
   Answer: D. 1·2·3 = 6 and 1·2·3·4/4 = 6. (Key point: Basis)

2. The next term added is?
   A. k(k+1)(k+2)
   B. (k+2)(k+3)(k+4)
   C. (k+1)(k+2)
   D. (k+1)(k+2)(k+3)
   Answer: D. Term k+1 is (k+1)(k+2)(k+3). (Key point: Term added)

3. The common factor taken out is?
   A. k(k+1)
   B. (k+3)
   C. (k+4)
   D. (k+1)(k+2)(k+3)
   Answer: D. Both terms contain (k+1)(k+2)(k+3). (Key point: Common factor)

4. k/4 + 1 equals?
   A. (k+1)/4
   B. (k+2)/4
   C. k/4
   D. (k+4)/4
   Answer: D. k/4 + 4/4 = (k+4)/4, which gives the RHS. (Key point: Bracket)

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