---
title: "Induction: 3+3·5+3·5²+...+3·5^n"
subject: "Discrete Structure"
chapter: 1
key: "semester_4/discrete_structure:induction-sum-3-times-5-powers"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/induction-sum-3-times-5-powers
site: BCT Engineering
---

# Induction: 3+3·5+3·5²+...+3·5^n

Discrete Structure · Chapter 1: Logic, Induction and Reasoning

- Asked in: 2072 Asoj, 6 marks; 2068 Bhadra, 8 marks (asked 2 times)
- Video: [MATHEMATICAL INDUCTION. 1+3+5+7+....+(2n-1)=n^2](https://www.youtube.com/watch?v=e3yvkCf1300) — Cornelius Hachibamba
- 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-3-times-5-powers
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-1

## Question

Use mathematical induction to prove that 3 + 3·5 + 3·5² + ...... + 3·5^n = 3(5^(n+1) − 1)/4 whenever n is a nonnegative integer.

## Short answer

Base n=0: LHS=3, RHS=3(5−1)/4=3. Assume true for k. For k+1, sum = 3(5^(k+1)−1)/4 + 3·5^(k+1) = 3(5^(k+2)−1)/4. By induction, equality holds for all n≥0.

## Answer

**To prove.** P(n): 3 + 3·5 + 3·5² + … + 3·5ⁿ = 3(5ⁿ⁺¹ − 1)/4 for every integer n ≥ 0.

#### Basis step (n = 0)
- LHS = 3·5⁰ = 3.
- RHS = 3(5¹ − 1)/4 = 3·4/4 = 3.
- LHS = RHS, so P(0) is true.

#### Inductive hypothesis
Assume P(k) is true for some k ≥ 0:

3 + 3·5 + … + 3·5ᵏ = 3(5ᵏ⁺¹ − 1)/4

#### Inductive step
We must show P(k+1): 3 + 3·5 + … + 3·5ᵏ + 3·5ᵏ⁺¹ = 3(5ᵏ⁺² − 1)/4.

```text
LHS = [3 + 3·5 + … + 3·5ᵏ] + 3·5ᵏ⁺¹
    = 3(5ᵏ⁺¹ − 1)/4 + 3·5ᵏ⁺¹          (inductive hypothesis)
    = 3(5ᵏ⁺¹ − 1)/4 + 12·5ᵏ⁺¹/4       (write 3·5ᵏ⁺¹ over 4)
    = 3(5ᵏ⁺¹ − 1 + 4·5ᵏ⁺¹)/4
    = 3(5·5ᵏ⁺¹ − 1)/4
    = 3(5ᵏ⁺² − 1)/4  = RHS
```

So P(k) → P(k+1) is true.

#### Conclusion
By the principle of mathematical induction, P(n) is true for every nonnegative integer n. ∎

## Check yourself

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

1. What is checked in the basis step for n ≥ 0?
   A. n = 1: LHS 3 + 15 = 18 against RHS
   B. n = 0: LHS = 3 and RHS = 3(5 − 1)/4 = 3
   C. n = k+1
   D. That the formula holds for all n
   Answer: B. For n = 0 the sum is just 3·5⁰ = 3, and 3(5¹ − 1)/4 = 3. (Key point: Basis step)

2. What is assumed in the inductive hypothesis?
   A. P(k+1) is true
   B. P(k): 3 + 3·5 + … + 3·5ᵏ = 3(5ᵏ⁺¹ − 1)/4
   C. P(n) for all n
   D. The formula is false for some k
   Answer: B. We assume the formula holds for some k ≥ 0, then show it for k+1. (Key point: Inductive hypothesis)

3. In the inductive step, what is added to the hypothesis sum to reach P(k+1)?
   A. 3·5ᵏ
   B. 3·5ᵏ⁺¹
   C. 3·5ᵏ⁺²
   D. 5ᵏ⁺¹
   Answer: B. The (k+1)th sum adds one more term, 3·5ᵏ⁺¹. (Key point: Adding the next term)

4. Combining 3(5ᵏ⁺¹ − 1)/4 + 3·5ᵏ⁺¹ gives which result?
   A. 3(5ᵏ⁺¹ − 1)/4
   B. 3(5ᵏ⁺² − 1)/4
   C. 3(2·5ᵏ⁺¹ − 1)/4
   D. 3(5ᵏ⁺³ − 1)/4
   Answer: B. 3(5ᵏ⁺¹ − 1 + 4·5ᵏ⁺¹)/4 = 3(5·5ᵏ⁺¹ − 1)/4 = 3(5ᵏ⁺² − 1)/4. (Key point: Key algebra)

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