Prove by induction: 1 + 2 + 4 + ×s + 2^n - 1 = 2^n - 1

Base case n = 1: the left side is 1 and the right side is 2^1 - 1 = 1.; Assume 1 + 2 + 4 + ×s + 2^k - 1 = 2^k - 1.; Step: 1 + 2 + 4 + ×s + 2^k - 1 + 2^k; = 2^k - 1 + 2^k; = 2 × 2^k - 1; = 2^k + 1 - 1; which is the formula for n = k + 1, so it holds for every n ≥ 1.

Prove by induction that 3 divides n^3 + 14n for every n ≥ 1

Base case n = 1: 1 + 14 = 15, which is divisible by 3.; Assume 3 divides k^3 + 14k.; Step: (k + 1)^3 + 14(k + 1); = k^3 + 3k^2 + 17k + 15; = (k^3 + 14k) + 3(k^2 + k + 5); Both parts of (k^3 + 14k) + 3(k^2 + k + 5) are divisible by 3, so 3 divides (k + 1)^3 + 14(k + 1).

Prove by induction: 10 + 50 + 250 + ×s + 10 × 5^n - 1 = (5(5^n - 1))/(2)

Base case n = 1: the left side is 10 and the right side is (5(5 - 1))/(2) = 10.; Assume 10 + 50 + 250 + ×s + 10 × 5^k - 1 = (5(5^k - 1))/(2).; Step: 10 + 50 + 250 + ×s + 10 × 5^k - 1 + 10 × 5^k; = (5(5^k - 1))/(2) + 10 × 5^k; = (25 × 5^k - 5)/(2); = (5(5^k + 1 - 1))/(2); which is the formula for n = k + 1, so it holds for every n ≥ 1.

Prove by induction: 5 + 30 + 180 + ×s + 5 × 6^n - 1 = 6^n - 1

Base case n = 1: the left side is 5 and the right side is 6 - 1 = 5.; Assume 5 + 30 + 180 + ×s + 5 × 6^k - 1 = 6^k - 1.; Step: 5 + 30 + 180 + ×s + 5 × 6^k - 1 + 5 × 6^k; = 6^k - 1 + 5 × 6^k; = 6 × 6^k - 1; = 6^k + 1 - 1; which is the formula for n = k + 1, so it holds for every n ≥ 1.

Write the statement for n = k + 1: 10 + 60 + 360 + ×s + 10 × 6^n - 1 = 2(6^n - 1)

10 + 60 + 360 + ×s + 10 × 6^k - 1 + 10 × 6^k = 2(6^k + 1 - 1)

Write the statement for n = k + 1: 7 + 49 + 343 + ×s + 7 × 7^n - 1 = (7(7^n - 1))/(6)

7 + 49 + 343 + ×s + 7 × 7^k - 1 + 7 × 7^k = (7(7^k + 1 - 1))/(6)

Write the statement for n = k + 1: 1 × 16 + 2 × 17 + 3 × 18 + ×s + n(n + 15) = (n(n + 1)(n + 23))/(3)

1 × 16 + 2 × 17 + 3 × 18 + ×s + k(k + 15) + (k + 1)(k + 16) = ((k + 1)(k + 2)(k + 24))/(3)

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: 9 + 14 + 19 + ×s + (5n + 4) = (n(5n + 13))/(2)

Base case n = 1: the left side is 9 and the right side is (1(18))/(2) = 9.; Assume 9 + 14 + 19 + ×s + (5k + 4) = (k(5k + 13))/(2).; Step: 9 + 14 + 19 + ×s + (5k + 4) + (5k + 9); = (k(5k + 13))/(2) + (5k + 9); = (5k^2 + 23k + 18)/(2); = ((k + 1)(5k + 18))/(2); which is the formula for n = k + 1, so it holds for every n ≥ 1.

Write the statement for n = k + 1: 7 + 12 + 17 + ×s + (5n + 2) = (n(5n + 9))/(2)

7 + 12 + 17 + ×s + (5k + 2) + (5k + 7) = ((k + 1)(5k + 14))/(2)

For P(n): 4 divides 13^n + 3, write P(1), P(k) and P(k + 1).

P(1): 4 divides 13^1 + 3; P(k): 4 divides 13^k + 3; P(k + 1): 4 divides 13^k + 1 + 3

Prove by induction that 4 divides 13^n + 3 for every integer n ≥ 1.

Base case n = 1: 13 + 3 = 16 = 4 × 4.; Assume P(k): 4 divides 13^k + 3.; Step: 13^k+1 + 3 = 13(13^k + 3) - 36; 4 divides 13^k + 3 (the assumption) and divides 36,; so P(k + 1) holds.

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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Sigma Prep · sigmaprep.io/worksheets Name Mathematical Induction Date Period Answer each question. Prove by induction Prove by induction: $1 + 2 + 4 + \dots + 2^{n - 1} = 2^{n} - 1$ Prove by induction that $3$ divides $n^{3} + 14 n$ for every $n \geq 1$ Prove by induction: $10 +$ $50 +$ $250 +$ $\dots +$ $10 \cdot 5^{n - 1} = \frac{5 (5^{n} - 1)}{2}$ Prove by induction: $5 + 30 + 180 + \dots + 5 \cdot 6^{n - 1} = 6^{n} - 1$ Write the statement for $n = k + 1$ Write the statement for $n = k + 1$ : $10 + 60 + 360 + \dots + 10 \cdot 6^{n - 1} = 2 (6^{n} - 1)$ Write the statement for $n = k + 1$ : $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{n - 1} = \frac{7 (7^{n} - 1)}{6}$ Write the statement for $n = k + 1$ : $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $n (n + 15) = \frac{n (n + 1) (n + 23)}{3}$ Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ 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
Prove by induction:
$1 + 2 + 4 + \dots + 2^{n - 1} = 2^{n} - 1$
Prove by induction that $3$ divides $n^{3} + 14 n$ for every $n \geq 1$
Prove by induction:
$10 +$ $50 +$ $250 +$ $\dots +$ $10 \cdot 5^{n - 1} = \frac{5 (5^{n} - 1)}{2}$
Prove by induction:
$5 + 30 + 180 + \dots + 5 \cdot 6^{n - 1} = 6^{n} - 1$

Write the statement for $n = k + 1$
Write the statement for $n = k + 1$ : $10 + 60 + 360 + \dots + 10 \cdot 6^{n - 1} = 2 (6^{n} - 1)$
Write the statement for $n = k + 1$ : $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{n - 1} = \frac{7 (7^{n} - 1)}{6}$
Write the statement for $n = k + 1$ : $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $n (n + 15) = \frac{n (n + 1) (n + 23)}{3}$
Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$



Sigma Prep · sigmaprep.io/worksheets Name Mathematical InductionAnswer keyVersion 1 Date Period Answer each question. Prove by induction Prove by induction: $1 + 2 + 4 + \dots + 2^{n - 1} = 2^{n} - 1$ $\begin{aligned} Base case n = 1 : the left side is 1 and the right side is 2^{1} - 1 = 1 . \\ Assume 1 + 2 + 4 + \dots + 2^{k - 1} = 2^{k} - 1 . \\ Step: 1 + 2 + 4 + \dots + 2^{k - 1} + 2^{k} \\ = 2^{k} - 1 + 2^{k} \\ = 2 \cdot 2^{k} - 1 \\ = 2^{k + 1} - 1 \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$ Prove by induction that $3$ divides $n^{3} + 14 n$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 1 + 14 = 15, which is divisible by 3 . \\ Assume 3 divides k^{3} + 14 k . \\ Step: {(k + 1)}^{3} + 14 (k + 1) \\ = k^{3} + 3 k^{2} + 17 k + 15 \\ = (k^{3} + 14 k) + 3 (k^{2} + k + 5) \\ Both parts of (k^{3} + 14 k) + 3 (k^{2} + k + 5) are divisible by 3, so 3 divides {(k + 1)}^{3} + 14 (k + 1) . \end{aligned}$ Prove by induction: $10 +$ $50 +$ $250 +$ $\dots +$ $10 \cdot 5^{n - 1} = \frac{5 (5^{n} - 1)}{2}$ $\begin{aligned} Base case n = 1 : the left side is 10 and the right side is \frac{5 (5 - 1)}{2} = 10 . \\ Assume 10 + 50 + 250 + \dots + 10 \cdot 5^{k - 1} = \frac{5 (5^{k} - 1)}{2} . \\ Step: 10 + 50 + 250 + \dots + 10 \cdot 5^{k - 1} + 10 \cdot 5^{k} \\ = \frac{5 (5^{k} - 1)}{2} + 10 \cdot 5^{k} \\ = \frac{25 \cdot 5^{k} - 5}{2} \\ = \frac{5 (5^{k + 1} - 1)}{2} \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$ Prove by induction: $5 + 30 + 180 + \dots + 5 \cdot 6^{n - 1} = 6^{n} - 1$ $\begin{aligned} Base case n = 1 : the left side is 5 and the right side is 6 - 1 = 5 . \\ Assume 5 + 30 + 180 + \dots + 5 \cdot 6^{k - 1} = 6^{k} - 1 . \\ Step: 5 + 30 + 180 + \dots + 5 \cdot 6^{k - 1} + 5 \cdot 6^{k} \\ = 6^{k} - 1 + 5 \cdot 6^{k} \\ = 6 \cdot 6^{k} - 1 \\ = 6^{k + 1} - 1 \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$ Write the statement for $n = k + 1$ Write the statement for $n = k + 1$ : $10 + 60 + 360 + \dots + 10 \cdot 6^{n - 1} = 2 (6^{n} - 1)$ $10 +$ $60 +$ $360 +$ $\dots +$ $10 \cdot 6^{k - 1} +$ $10 \cdot 6^{k} = 2 (6^{k + 1} - 1)$ Write the statement for $n = k + 1$ : $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{n - 1} = \frac{7 (7^{n} - 1)}{6}$ $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{k - 1} +$ $7 \cdot 7^{k} = \frac{7 (7^{k + 1} - 1)}{6}$ Write the statement for $n = k + 1$ : $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $n (n + 15) = \frac{n (n + 1) (n + 23)}{3}$ $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $k (k + 15) +$ $(k + 1) (k + 16) = \frac{(k + 1) (k + 2) (k + 24)}{3}$ Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ $2 +$ $6 +$ $18 +$ $\dots +$ $2 \cdot 3^{k - 1} +$ $2 \cdot 3^{k} = 3^{k + 1} -$ $1$ 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
Prove by induction:
$1 + 2 + 4 + \dots + 2^{n - 1} = 2^{n} - 1$ $\begin{aligned} Base case n = 1 : the left side is 1 and the right side is 2^{1} - 1 = 1 . \\ Assume 1 + 2 + 4 + \dots + 2^{k - 1} = 2^{k} - 1 . \\ Step: 1 + 2 + 4 + \dots + 2^{k - 1} + 2^{k} \\ = 2^{k} - 1 + 2^{k} \\ = 2 \cdot 2^{k} - 1 \\ = 2^{k + 1} - 1 \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$
Prove by induction that $3$ divides $n^{3} + 14 n$ for every $n \geq 1$ $\begin{aligned} Base case n = 1 : 1 + 14 = 15, which is divisible by 3 . \\ Assume 3 divides k^{3} + 14 k . \\ Step: {(k + 1)}^{3} + 14 (k + 1) \\ = k^{3} + 3 k^{2} + 17 k + 15 \\ = (k^{3} + 14 k) + 3 (k^{2} + k + 5) \\ Both parts of (k^{3} + 14 k) + 3 (k^{2} + k + 5) are divisible by 3, so 3 divides {(k + 1)}^{3} + 14 (k + 1) . \end{aligned}$
Prove by induction:
$10 +$ $50 +$ $250 +$ $\dots +$ $10 \cdot 5^{n - 1} = \frac{5 (5^{n} - 1)}{2}$ $\begin{aligned} Base case n = 1 : the left side is 10 and the right side is \frac{5 (5 - 1)}{2} = 10 . \\ Assume 10 + 50 + 250 + \dots + 10 \cdot 5^{k - 1} = \frac{5 (5^{k} - 1)}{2} . \\ Step: 10 + 50 + 250 + \dots + 10 \cdot 5^{k - 1} + 10 \cdot 5^{k} \\ = \frac{5 (5^{k} - 1)}{2} + 10 \cdot 5^{k} \\ = \frac{25 \cdot 5^{k} - 5}{2} \\ = \frac{5 (5^{k + 1} - 1)}{2} \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$
Prove by induction:
$5 + 30 + 180 + \dots + 5 \cdot 6^{n - 1} = 6^{n} - 1$ $\begin{aligned} Base case n = 1 : the left side is 5 and the right side is 6 - 1 = 5 . \\ Assume 5 + 30 + 180 + \dots + 5 \cdot 6^{k - 1} = 6^{k} - 1 . \\ Step: 5 + 30 + 180 + \dots + 5 \cdot 6^{k - 1} + 5 \cdot 6^{k} \\ = 6^{k} - 1 + 5 \cdot 6^{k} \\ = 6 \cdot 6^{k} - 1 \\ = 6^{k + 1} - 1 \\ which is the formula for n = k + 1, so it holds for every n \geq 1 . \end{aligned}$

Write the statement for $n = k + 1$
Write the statement for $n = k + 1$ : $10 + 60 + 360 + \dots + 10 \cdot 6^{n - 1} = 2 (6^{n} - 1)$ $10 +$ $60 +$ $360 +$ $\dots +$ $10 \cdot 6^{k - 1} +$ $10 \cdot 6^{k} = 2 (6^{k + 1} - 1)$
Write the statement for $n = k + 1$ : $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{n - 1} = \frac{7 (7^{n} - 1)}{6}$ $7 +$ $49 +$ $343 +$ $\dots +$ $7 \cdot 7^{k - 1} +$ $7 \cdot 7^{k} = \frac{7 (7^{k + 1} - 1)}{6}$
Write the statement for $n = k + 1$ : $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $n (n + 15) = \frac{n (n + 1) (n + 23)}{3}$ $1 \cdot 16 +$ $2 \cdot 17 +$ $3 \cdot 18 +$ $\dots +$ $k (k + 15) +$ $(k + 1) (k + 16) = \frac{(k + 1) (k + 2) (k + 24)}{3}$
Write the statement for $n = k + 1$ : $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ $2 +$ $6 +$ $18 +$ $\dots +$ $2 \cdot 3^{k - 1} +$ $2 \cdot 3^{k} = 3^{k + 1} -$ $1$

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 $1 + 3 + 5 + \dots + (2 k - 1) + (2 k + 1) = {(k + 1)}^{2}$ .

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:
$9 + 14 + 19 + \dots + (5 n + 4) = \frac{n (5 n + 13)}{2}$
→ Answer
$Base case n = 1$ : the left side is $9$ and the right side is $\frac{1 (18)}{2} = 9$ .
$Assume 9 +$ $14 +$ $19 +$ $\dots +$ $(5 k + 4) = \frac{k (5 k + 13)}{2}$ .
$Step: 9 +$ $14 +$ $19 +$ $\dots +$ $(5 k + 4) +$ $(5 k + 9)$
$= \frac{k (5 k + 13)}{2} + (5 k + 9)$
$= \frac{5 k^{2} + 23 k + 18}{2}$
$= \frac{(k + 1) (5 k + 18)}{2}$
which is the formula for $n = k + 1$ , so it holds for every $n \geq 1$ .
Write the statement for n = k + 1: Write the statement for $n = k + 1$ : $7 + 12 + 17 + \dots + (5 n + 2) = \frac{n (5 n + 9)}{2}$ → $7 +$ $12 +$ $17 +$ $\dots +$ $(5 k + 2) +$ $(5 k + 7) = \frac{(k + 1) (5 k + 14)}{2}$
Write P(1), P(k) and P(k + 1): For $P (n)$ : $4$ divides $13^{n} + 3$ , write $P (1)$ , $P (k)$ and $P (k + 1)$ . → $P (1)$ : $4$ divides $13^{1} + 3$
$P (k)$ : $4$ divides $13^{k} + 3$
$P (k + 1)$ : $4$ divides $13^{k + 1} + 3$
Divisibility and inequalities: Prove by induction that $4$ divides $13^{n} + 3$ for every integer $n \geq 1$ .
→ Answer
$Base case n = 1$ : $13 + 3 = 16 = 4 \cdot 4$ .
$Assume P (k)$ : $4$ divides $13^{k} + 3$ .
$Step: 13^{k + 1} + 3 = 13 ⁣ (13^{k} + 3) - 36$
4 divides $13^{k} + 3$ (the assumption) and divides $36$ ,
so $P (k + 1)$ holds.

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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.

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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 4: prove by induction, write the statement for n = k + 1, write p(1), p(k) and p(k + 1) and divisibility and inequalities. Tick as many as you want and set how many of each, or let it spread them evenly.

Proof by mathematical induction worksheet

Select your difficulty

Prove by induction

Write the statement for n=k+1n = k + 1n=k+1

Prove by induction

Write the statement for n=k+1n = k + 1n=k+1

How to do these

Where students go wrong

What you can put on this worksheet

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Write the statement for $n = k + 1$

Write the statement for $n = k + 1$