Prove by induction that 3 divides 4^n - 1 for every n ≥ 1

Base case n = 1: 4 - 1 = 3, which is divisible by 3.; Assume 3 divides 4^k - 1.; Step: 4^k + 1 - 1; = 4 × 4^k - 4 + 3; = 4(4^k - 1) + 3; Both parts of 4(4^k - 1) + 3 are divisible by 3, so 3 divides 4^k + 1 - 1.

Write the statement for n = k + 1: 2 + 6 + 18 + ×s + 2 × 3^n - 1 = 3^n - 1

2 + 6 + 18 + ×s + 2 × 3^k - 1 + 2 × 3^k = 3^k + 1 - 1

Prove by induction that 3 divides n^3 + 2n for every n ≥ 1

Base case n = 1: 1 + 2 = 3, which is divisible by 3.; Assume 3 divides k^3 + 2k.; Step: (k + 1)^3 + 2(k + 1); = k^3 + 3k^2 + 5k + 3; = (k^3 + 2k) + 3(k^2 + k + 1); Both parts of (k^3 + 2k) + 3(k^2 + k + 1) are divisible by 3, so 3 divides (k + 1)^3 + 2(k + 1).

Write the statement for n = k + 1: 1^2 + 2^2 + 3^2 + ×s + n^2 = (n(n + 1)(2n + 1))/(6)

1^2 + 2^2 + 3^2 + ×s + k^2 + (k + 1)^2 = ((k + 1)(k + 2)(2k + 3))/(6)

Prove by induction that 5 divides 6^n - 1 for every n ≥ 1

Base case n = 1: 6 - 1 = 5, which is divisible by 5.; Assume 5 divides 6^k - 1.; Step: 6^k + 1 - 1; = 6 × 6^k - 6 + 5; = 6(6^k - 1) + 5; Both parts of 6(6^k - 1) + 5 are divisible by 5, so 5 divides 6^k + 1 - 1.

Write the statement for n = k + 1: 1 × 2 + 2 × 3 + 3 × 4 + ×s + n(n + 1) = (n(n + 1)(n + 2))/(3)

1 × 2 + 2 × 3 + 3 × 4 + ×s + k(k + 1) + (k + 1)(k + 2) = ((k + 1)(k + 2)(k + 3))/(3)

Prove by induction that 4 divides 5^n - 1 for every n ≥ 1

Base case n = 1: 5 - 1 = 4, which is divisible by 4.; Assume 4 divides 5^k - 1.; Step: 5^k + 1 - 1; = 5 × 5^k - 5 + 4; = 5(5^k - 1) + 4; Both parts of 5(5^k - 1) + 4 are divisible by 4, so 4 divides 5^k + 1 - 1.

Write the statement for n = k + 1: 1 + 3 + 9 + ×s + 3^n - 1 = (3^n - 1)/(2)

1 + 3 + 9 + ×s + 3^k - 1 + 3^k = (3^k + 1 - 1)/(2)

Prove by induction: 2 + 4 + 6 + ×s + 2n = n(n + 1)

Base case n = 1: the left side is 2 and the right side is 1(2) = 2.; Assume 2 + 4 + 6 + ×s + 2k = k(k + 1).; Step: 2 + 4 + 6 + ×s + 2k + 2(k + 1); = k(k + 1) + 2(k + 1); = (k + 1)(k + 2); extwhich is the formula for n = k + 1 ext, so it holds for every n ≥ 1 ext.

Write the statement for n = k + 1: 1 + 2 + 4 + ×s + 2^n - 1 = 2^n - 1

1 + 2 + 4 + ×s + 2^k - 1 + 2^k = 2^k + 1 - 1

Worksheets · Precalculus

Proof by mathematical induction worksheet

Three parts. Check the statement for n = 1. Assume it is true for some n = k. Then use that assumption to show it is true for n = k + 1. For a sum, the step is to take the formula for k and add the next term; for divisibility, it is to rewrite the k + 1 case so the k case appears inside it.

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Pick more than one for a sheet that mixes them.



Sigma Prep · sigmaprep.io/worksheets Name Mathematical Induction Date Period Answer each question. Prove by induction that $3$ divides $4^{n} - 1$ for every $n \geq 1$ Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ Prove by induction that $3$ divides $n^{3} + 2 n$ for every $n \geq 1$ Write the statement for $n = k + 1$ : $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ Prove by induction that $5$ divides $6^{n} - 1$ for every $n \geq 1$ Write the statement for $n = k + 1$ : $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n (n + 1) (n + 2)}{3}$ Prove by induction that $4$ divides $5^{n} - 1$ for every $n \geq 1$ Write the statement for $n = k + 1$ : $1 + 3 + 9 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Mathematical Induction

Date Period

Answer each question.

Prove by induction that $3$ divides $4^{n} - 1$ for every $n \geq 1$
Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$
Prove by induction that $3$ divides $n^{3} + 2 n$ for every $n \geq 1$
Write the statement for $n = k + 1$ : $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
Prove by induction that $5$ divides $6^{n} - 1$ for every $n \geq 1$
Write the statement for $n = k + 1$ : $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n (n + 1) (n + 2)}{3}$
Prove by induction that $4$ divides $5^{n} - 1$ for every $n \geq 1$
Write the statement for $n = k + 1$ : $1 + 3 + 9 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$



Sigma Prep · sigmaprep.io/worksheets Name Mathematical InductionAnswer keyVersion 1 Date Period Answer each question. Prove by induction that $3$ divides $4^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 4 - 1 = 3, which is divisible by 3 . \\ Assume 3 divides 4^{k} - 1 . \\ Step: 4^{k + 1} - 1 \\ = 4 \cdot 4^{k} - 4 + 3 \\ = 4 (4^{k} - 1) + 3 \\ Both parts of 4 (4^{k} - 1) + 3 are divisible by 3, so 3 divides 4^{k + 1} - 1 . \end{aligned}$ Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ $2 + 6 + 18 + \dots + 2 \cdot 3^{k - 1} + 2 \cdot 3^{k} = 3^{k + 1} - 1$ Prove by induction that $3$ divides $n^{3} + 2 n$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 1 + 2 = 3, which is divisible by 3 . \\ Assume 3 divides k^{3} + 2 k . \\ Step: (k + 1)^{3} + 2 (k + 1) \\ = k^{3} + 3 k^{2} + 5 k + 3 \\ = (k^{3} + 2 k) + 3 (k^{2} + k + 1) \\ Both parts of (k^{3} + 2 k) + 3 (k^{2} + k + 1) are divisible by 3, so 3 divides (k + 1)^{3} + 2 (k + 1) . \end{aligned}$ Write the statement for $n = k + 1$ : $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} = \frac{(k + 1) (k + 2) (2 k + 3)}{6}$ Prove by induction that $5$ divides $6^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 6 - 1 = 5, which is divisible by 5 . \\ Assume 5 divides 6^{k} - 1 . \\ Step: 6^{k + 1} - 1 \\ = 6 \cdot 6^{k} - 6 + 5 \\ = 6 (6^{k} - 1) + 5 \\ Both parts of 6 (6^{k} - 1) + 5 are divisible by 5, so 5 divides 6^{k + 1} - 1 . \end{aligned}$ Write the statement for $n = k + 1$ : $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n (n + 1) (n + 2)}{3}$ $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k (k + 1) + (k + 1) (k + 2) = \frac{(k + 1) (k + 2) (k + 3)}{3}$ Prove by induction that $4$ divides $5^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 5 - 1 = 4, which is divisible by 4 . \\ Assume 4 divides 5^{k} - 1 . \\ Step: 5^{k + 1} - 1 \\ = 5 \cdot 5^{k} - 5 + 4 \\ = 5 (5^{k} - 1) + 4 \\ Both parts of 5 (5^{k} - 1) + 4 are divisible by 4, so 4 divides 5^{k + 1} - 1 . \end{aligned}$ Write the statement for $n = k + 1$ : $1 + 3 + 9 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ $1 + 3 + 9 + \dots + 3^{k - 1} + 3^{k} = \frac{3^{k + 1} - 1}{2}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Mathematical InductionAnswer keyVersion 1

Date Period

Answer each question.

Prove by induction that $3$ divides $4^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 4 - 1 = 3, which is divisible by 3 . \\ Assume 3 divides 4^{k} - 1 . \\ Step: 4^{k + 1} - 1 \\ = 4 \cdot 4^{k} - 4 + 3 \\ = 4 (4^{k} - 1) + 3 \\ Both parts of 4 (4^{k} - 1) + 3 are divisible by 3, so 3 divides 4^{k + 1} - 1 . \end{aligned}$
Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ $2 + 6 + 18 + \dots + 2 \cdot 3^{k - 1} + 2 \cdot 3^{k} = 3^{k + 1} - 1$
Prove by induction that $3$ divides $n^{3} + 2 n$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 1 + 2 = 3, which is divisible by 3 . \\ Assume 3 divides k^{3} + 2 k . \\ Step: (k + 1)^{3} + 2 (k + 1) \\ = k^{3} + 3 k^{2} + 5 k + 3 \\ = (k^{3} + 2 k) + 3 (k^{2} + k + 1) \\ Both parts of (k^{3} + 2 k) + 3 (k^{2} + k + 1) are divisible by 3, so 3 divides (k + 1)^{3} + 2 (k + 1) . \end{aligned}$
Write the statement for $n = k + 1$ : $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} = \frac{(k + 1) (k + 2) (2 k + 3)}{6}$
Prove by induction that $5$ divides $6^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 6 - 1 = 5, which is divisible by 5 . \\ Assume 5 divides 6^{k} - 1 . \\ Step: 6^{k + 1} - 1 \\ = 6 \cdot 6^{k} - 6 + 5 \\ = 6 (6^{k} - 1) + 5 \\ Both parts of 6 (6^{k} - 1) + 5 are divisible by 5, so 5 divides 6^{k + 1} - 1 . \end{aligned}$
Write the statement for $n = k + 1$ : $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n (n + 1) (n + 2)}{3}$ $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k (k + 1) + (k + 1) (k + 2) = \frac{(k + 1) (k + 2) (k + 3)}{3}$
Prove by induction that $4$ divides $5^{n} - 1$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 5 - 1 = 4, which is divisible by 4 . \\ Assume 4 divides 5^{k} - 1 . \\ Step: 5^{k + 1} - 1 \\ = 5 \cdot 5^{k} - 5 + 4 \\ = 5 (5^{k} - 1) + 4 \\ Both parts of 5 (5^{k} - 1) + 4 are divisible by 4, so 4 divides 5^{k + 1} - 1 . \end{aligned}$
Write the statement for $n = k + 1$ : $1 + 3 + 9 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ $1 + 3 + 9 + \dots + 3^{k - 1} + 3^{k} = \frac{3^{k + 1} - 1}{2}$

How to do these

Prove by induction: $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$

Base case: for n = 1 both sides are 1.
Assume 1 + 3 + ... + (2k - 1) = k squared.
Add the next odd number, 2k + 1, to both sides: the left is the sum up to k + 1, and the right is k squared + 2k + 1, which is (k + 1) squared.

The answer is $(k + 1)^{2}$ .

Where students go wrong

Proving the k + 1 case without using the assumption, or assuming what is to be proved. The step has to start from the k case and reach the k + 1 case; if the assumption never appears in the algebra, it was not an induction proof.

Also called proof by induction, induction proofs, principle of mathematical induction or inductive step.

What you can put on this worksheet

Prove by induction: Prove by induction: $2 + 4 + 6 + \dots + 2 n = n (n + 1)$ → $\begin{aligned} Base case n = 1 : the left side is 2 and the right side is 1 (2) = 2 . \\ Assume 2 + 4 + 6 + \dots + 2 k = k (k + 1) . \\ Step: 2 + 4 + 6 + \dots + 2 k + 2 (k + 1) \\ = k (k + 1) + 2 (k + 1) \\ = (k + 1) (k + 2) \\ e x t w h i c h i s t h e f o r m u l a f o r n = k + 1 e x t, s o i t h o l d s f o r e v e r y n \geq 1 e x t . \end{aligned}$
Write the statement for n = k + 1: Write the statement for $n = k + 1$ : $1 + 2 + 4 + \dots + 2^{n - 1} = 2^{n} - 1$ → $1 + 2 + 4 + \dots + 2^{k - 1} + 2^{k} = 2^{k + 1} - 1$

Questions about these worksheets

Yes. Every proof by mathematical induction sheet is free to print and free to download as a PDF. There is no account to make, no email to hand over and no limit on how many you take.

Yes. The answer key prints on a second page, and you can choose whether it shows each question beside its answer or just the answers in a list.

Yes. Print straight from the page, or download the PDF and print that. The sheet is laid out for paper rather than squeezed off a screen, so there is room to work under each question, and the answer key comes out on its own page.

Precalculus is usually taken around 11th or 12th grade, though schools vary and plenty of students meet it earlier or later. Pick the difficulty rather than the grade: the easy level suits a first lesson on it, the hard level suits review before a test.

Yes, as many as you like. The questions are built fresh each time rather than picked from a fixed set of files, so pressing Generate gives a new sheet. That is what you want for a second class, a retake, or two students sitting next to each other.

That is the closest comparison, and Kuta Software is good software. It is also paid software you install on a computer. These worksheets run in a browser tab for free, with no account and nothing to install. Like Kuta, the proof by mathematical induction questions are generated when you ask for them rather than pulled from a fixed set of files, so two classes never get the same sheet, and the answer key comes with it.

Yes. There are three levels, and you can tick more than one for a sheet that mixes them, in which case the questions come out easiest first. The harder levels are not just larger numbers: they ask for something the easy ones do not.

You pick from 2: prove by induction and write the statement for n = k + 1. Tick as many as you want and set how many of each, or let it spread them evenly.

Proof by mathematical induction worksheet

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Mathematical Induction

Mathematical InductionAnswer keyVersion 1

How to do these

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