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102 LearnersLast updated on September 29, 2025
Mathematical Induction
Mathematical induction is a step-by-step proof method used to show that a statement is true for all natural numbers. It is especially useful for proving algebraic statements that involve a variable n, where n represents a natural number.
What is Mathematical Induction?
Mathematical induction is often applied to prove formulas involving natural numbers. For example, the formula to find the sum of the first n natural numbers is n (n + 1)2. Now, mathematical induction can be used to verify if the formula holds true for every natural number n.
Principle of Mathematical Induction Statement
The principle of mathematical induction is widely used to prove a result, and it is also a fundamental concept in mathematics. In the principle of mathematical induction, the statement to be proved is usually written as P(n), where n is a natural number. While using this, we need to understand some statements that help solve the result.
- Start by verifying the base case: show that the statement is true for n = 1, which means proving that P(1) is true.
- If the statement is true for n = k, where k is a natural number, we assume P(k) is true. This assumption is called the inductive hypothesis.
- Then, the inductive hypothesis is used to prove that n = k + 1, P(k + 1) is also true.
Principle of Mathematical Induction Proof
The principle of mathematical induction is a concept in mathematics used to prove statements that are true about natural numbers. There is a form that is used to solve the problem, the form is P(n), where n is all natural numbers. The principle has two types:
- Base step
- Inductive step
Base step:
The first part of a mathematical induction proof is called the base step. Here, we check whether the statement holds true for the smallest natural number, usually 1 or sometimes 0.
Inductive step:
Here, we assume that the statement is true for some natural number n = k. This assumption is called the inductive hypothesis. Then, we use this assumption to prove that the statement must also be true for n = k + 1.
If both the base step and inductive step work, then by mathematical induction, the statement is proven true for all natural numbers starting from the base case.
How to Solve Mathematical Induction?
- While solving mathematical induction, you need to check whether the statements are true or not for the natural number n = 1.
- Substitute the base value into both sides of the equation to verify if the statement holds true
- If the result is equal on both sides, then the statement is true.
- Then, assume the statement is true for some natural number n = k, write the form as P (k).
- Prove the statement is true for n = k + 1
- In the statement, replace the n with k + 1 in the formula. Try to simplify the statement and show that it matches what the formula says it should be.
For example,
Prove that 1 + 3 + 5 + … + (2n - 1) = n2
Solution:
The left-hand side represents the sum of the first n odd numbers. We will use mathematical induction to prove that this sum equals n2
The first step is the statement is true or not by using the base step
LHS = RHS
LHS = 1 = 1
RHS = n2 = 12 = 1
The first step is to prove that the statement is true by using the base step.
The next step is inductive, assuming that the formula is true for n = k
Which means, 1 + 3 + 5 + … + (2k -1) = k2
The final step is an inductive step to prove that the formula is true for n = k + 1
1 + 3 + 5 +. . . + (2k -1) + (2(k +1) -1) = (k + 1)2
= k2 + (2k + 1) = k2 + 2k + 1 = (k + 1)2
1 + 3 + 5 +. …+ (2n - 1) = n2 is proven to be true for all natural numbers n ≥ 1.
Real Life Applications of Mathematical Induction
Mathematical induction may seem like just a classroom concept, but it plays a role in many real-world and practical areas, especially in fields where logic, patterns are used.
- Computer Programming: Many programs involve loops and recursion, which are based on repeated rules. Programmers are proving that a function works for all inputs. They often use induction to prove that the function produces the correct result for any number of inputs.
- Robot Movement or Game Design: In robotics, mathematical induction is used for moving robots that depend on previous actions. The method helps prove that the robot will always reach its goal, as long as it starts correctly and follows each step in the sequence.
- Banking and Finance: In some financial systems that involve patterns like compound interest, EMI payments, or number patterns, we use mathematical induction. In finance, there is a formula that calculates the total payment over a period of n months. After that, bankers and data scientists might use induction to ensure it works for all periods.
- Construction: Engineers use mathematical induction for repeated design in elements like beams, tiles, etc. To show a pattern of bricks or tiles that will fit perfectly in every layer of a wall or floor, they use induction to prove the pattern works for all levels.
- Software Engineering: Software engineering uses mathematical induction for algorithm analysis. Induction helps prove an algorithm’s correctness and efficiency for any size input. When analyzing algorithms such as binary search or merge sort, computer scientists use induction to prove that they work no matter how many elements are in the list.
Common Mistakes and How to Avoid Them on Mathematical Induction
Mathematical induction is a logical method to prove that a result is true. Students may make small mistakes that lead to wrong results. Here are some mistakes that help to avoid while solving problems.
Mistake 1
Skipping or doing the base case incorrectly
Sometimes students skip the first step (Base step) and start doing the second step (inductive step) without checking the base step. It will give a wrong result. So, always start with the first step (Base step). Add the value in the n = 1 on both sides. Make sure that both sides are equal.
Mistake 2
Assuming without stating the inductive hypothesis
Students considered that the formula is true for n = k, but they didn’t prove that the statement is true clearly. Always write the inductive hypothesis like the statement is true for n = k, P(k).
Mistake 3
Wrongly applying the inductive hypothesis in step 3
Many students often use the assumption P(k), but they make mistakes when adding it in while trying to prove P(k + 1). Start with LHS for P(k + 1) = LHS for P(k) + next term.
Mistake 4
Forgetting to simplify completely
Some students leave the equation in half and simplify it. This will make it unclear whether P(k + 1) is true. Keep simplifying until the equation exactly matches the right-hand side of P(k + 1).
Mistake 5
Thinking that one example proves everything
Students tried that n = 1 + 2 + 3 + 4 and said that it will work for all elements, which is not proof. Mathematical induction is not guessing, by proving that logic. So always do the three steps: 1) Base step, 2) Inductive Hypothesis, and 3) Inductive steps.
Solved Examples of Mathematical Induction
Problem 1
Prove that 1 + 2 + 3 +. . . + n = n (n +1)2 for all natural number n 1
The formula is true for k + 1. This is true for all natural numbers n 1.
Explanation
Step 1:
Start with the Base step
Check for n = 1
LHS = 1
RHS = n (n +1)2 = 1 (1 +1)2 = 22 = 1
Then do step 2:
Inductive Hypothesis
Assume the statement is true for n = k
The statement is 1 + 2 + 3 +. . . + k = k (k +1)2
This is an assumption.
Step 3: Inductive step
Here, need to prove the statement for n = k + 1
1 + 2 + 3 +. . . + k + (k + 1) = k (k +1) (k + 2)2
Start from the LHS using the assumption:
(1 + 2 +. . . + k) + (k + 1) = k (k +1)2 + (k + 1)
Make a denominator in common:
= k (k +1) + 2(k + 1)2 = k (k +1) (k + 2)2
This is equal to the RHS.
The formula is true for k + 1. This is true for all natural numbers n 1.
Problem 2
Prove that 2n > n for all n 1
This is true for all k 1
Explanation
Step 1:
Base step n = 1
2n = 2 > 1
21 = 2 > 1
It is true
Step 2:
Inductive Hypothesis
Assume the statement is true for n = k
2n = 2 > 1
2k = k > 1
Step 3:
Inductive step
Here, need to prove that the statement for n = k + 1
2n = 2 > 1
2k + 1 > k + 1
Now multiple both sides using the inductive hypotheses
2k + 1 = 2(2k) + 2 (k)
Then need to show
2k k + 1
k 1
This is true for all k 1
Problem 3
Prove that 12 + 22 + 32 +... + n2 = n (n + 1)(2n + 1)6
This is true for all natural numbers n 1.
Explanation
Step 1:
Start with the Base step
Check for n = 1
LHS = 12 = 1
RHS = n (n +1)6 = 1 (1 +1) (2(1) + 1)6 = 1 × 2 × 36 = 1
LHS = RHS
Then do step 2:
Inductive Hypothesis
Assume the statement is true for n = k
The statement is 12 + 22 + 32 +. . . + k2 = k (k +1)(2k + 1)6
This is an assumption.
Step 3: Inductive step
Here, need to prove the statement for n = k + 1
12 + 22 + 32 +. . . + k2 + (k + 1)2 = k (k +1) (2k + 1)6
Start from LHS using the assumption:
(12 + 22 +. . . + k2) + (k + 1)2 = k (k +1)(2k + 1)6 + (k + 1)2
Factor (k + 1) from both terms:
= (k + 1) k(2k + 1)6 + (k + 1)
Put the denominator common for:
= (k + 1) k(2k + 1) + 6(k +1)6
= (k + 1) 2k2 + k + 6k +66
= (k + 1) 2k2 + 7k +66
Factor the quadratic:
= (k + 1) (k + 2) (2k + 3)6 = (k +1)(k + 2)(2k + 3)6
This is equal to the RHS.
The formula is true for k + 1. This is true for all natural numbers n 1.
Problem 4
Prove that 7n -1 is divisible by 6 for all n 1
7n−1 is divisible by 6 for all n 1
Explanation
Base step:
n = 1
7n -1
71 -1 = 6
It is true
Step 2:
Inductive Hypothesis
Assume the statement is true for n = k
That will come 7k -1 is divisible by 6
7k − 1 = 6m for some integer m
Step 3:
Inductive step
Here, need to prove that the statement for n = k + 1
7n -1
7k + 1 = 7 × 7k
7k +1 -1 = 7 × 7k - 1
=7 × 7k - 7 +7 - 1
= 7 (7k -1) + 6
By hypothesis, 7k -1 = 6m
7(6m) + 6 = 42m + 6 = 6(7m +1)
7n−1 is divisible by 6 for all n ≥ 1
Problem 5
Prove that 11n - 4n is divisible by 7 for all n ≥ 1
11n−4n is divisible by 7 for all n ≥ 1
Explanation
Base step:
n = 1
11n -41
11 -4 = 7
It is true
Step 2:
Inductive Hypothesis
Assume the statement is true for n = k
That will come 11k -4k is divisible by 7
11k − 4n = 7m for some integer m. This equation is divisible by 7
Step 3:
Inductive step
Here, need to prove that the statement for n = k + 1
11n -4n
11k + 1 -4k +1 = 11 × 11k - 4 × 4k
11 × 11k - 4 × 4k = 11(11k) -4(4k)
Group terms like this
= 11k(11) - 4k (4)
By hypothesis, 11k -4k = 7m
Try modulo 7:
11 ≡ 4 (mod7)
11n ≡ 4n (mod7)
11n−4n ≡ 0 (mod7)
11n−4n is divisible by 7 for all n ≥ 1
FAQs on Mathematical Induction
1.What does mathematical induction mean?
2.What is the principle of mathematical induction?
3.What are the steps involved in mathematical induction?
4.Why do we use the principle of mathematical induction?
5.Why do we assume that the statement is true for n=kn = kn=k in the inductive step?
6.How does learning Algebra help students in United States make better decisions in daily life?
7.How do technology and digital tools in United States support learning Algebra and Mathematical Induction ?
8.How can cultural or local activities in United States support learning Algebra topics such as Mathematical Induction ?
9.Does learning Algebra support future career opportunities for students in United States?
Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
