Principle of Mathematical Induction

Mathematical induction is a concept in mathematics that is used to prove various mathematical statements and theorems. The principle of mathematical induction is sometimes referred to as PMI. It is a technique used to prove basic theorems in mathematics that involve solving up to n finite natural terms.

 It is widely used in proving various statements such as a sum of first n natural numbers is given by the formula n(n + 1)/2.

This can be easily proved using the Principle of Mathematical Induction.

Suppose P(n) is a statement for n natural numbers, then it can be proved using the Principle of Mathematical Induction.

• We will prove for P(1), then let P(k) be true then prove for P(k+1).
• If P(k+1) holds, then we say that P(n) is true by the principle of mathematical induction.

We can compare mathematical induction to falling dominoes. When a domino falls, it knocks down the next domino in succession. The first domino knocks down the second one, the second one knocks down the third, and so on. In the end, all of the dominoes will be bowled over. But there are some conditions to be fulfilled:

• The base step is that the starting domino must fall to set the knocking process in action.
• The distance between dominoes must be equal for any two adjacent dominoes. Otherwise, a certain domino may fall without bowling over the next. Then the sequence of reactions will stop.
• Maintaining the equal inter-domino distance ensures that P(k) ⇒ P(k + 1) for each integer k ≥ a. This is the inductive step.

Principle of Mathematical Induction Statement

Any statement P(n), which is for "n" a natural number, can be proved using the Principle of Mathematical Induction by following the steps below,

Step 1: Verify if the statement is true for trivial cases (n = 1) i.e. check if P(1) is true.

Step 2: Assume that the statement is true for n = k for some k ≥ 1 i.e. P(k) is true.

Step 3: If the truth of P(k) implies the truth of P(k + 1), then the statement P(n) is true for all n ≥ 1.

The image added below contains all the steps of Mathematical Induction

The first statement is a fact, and if it is not possible for all P(n) to hold at n = 1, then these statements are true for some other values of n, say n = 2, n = 3, and others.

If the statement is true for P(k), then if P(k+1) is proven to be true, then we say that P(n) is true for all n belonging to Natural Numbers (N)

Mathematical Induction Solution and Proof

Consider a statement P(n), where n is a natural number. Then, to determine the validity of P(n) for every n, use the following principle:

• Base Step: Prove P(k) is true for k =1
• Assumption Step: Assume that the given statement P (n) is also true for n = k, whether k is any positive integer.
• Induction Step: Prove P(k+1) is true using basic mathematical properties.

If the above three steps are proved, then we can say that "By the principle of mathematical induction, P(n) is true for all n belonging to N".

Proof:

The principle starts with a base case, which is a statement confirmed to be true for a specific starting value. Next comes the inductive step, where we assume the statement holds for some positive integer k (this is the inductive hypothesis) and then show that it must also hold for k+1. If both parts are satisfied, the statement is proven true for all required values of n.

Mathematical Induction Example

Mathematical induction is used to prove various statements. We can learn this with the help of the following example.

Example: For any positive integer n, prove that n³ + 2n is always divisible by 3

Solution:

Let P(n): n³ + 2n is divisible by 3 be the given statement.

Step 1: Basic Step

Firstly we prove that P(1) is true. Let n = 1 in n³ + 2n
= 1³ + 2(1) 
= 3

As 3 is divisible by 3. Hence, P(1) is true.

Step 2: Assumption Step

Let us assume that P(k) is true
Then, k³ + 2k is divisible by 3
Thus, we can assume k³ + 2k = 3n,             (where n is any positive integer)....(i)

Step 3: Induction Steps

Now we have to prove that algebraic expression (k + 1)³ + 2(k + 1) is divisible by 3

= (k + 1)³ + 2(k + 1) 
= k³ + 3k² + 3k + 2k + 3
= (k³ + 2 k) + (3k² + 3k + 3)

from eq(i)

= 3n + 3(k² + k + 1)
= 3(n + k² + k + 1)

As it is a multiple of 3 we can say that it is divisible by 3.

Thus, P(k+1) is true i.e. (k + 1)³ + 2(k + 1) is be divisible by 3. Now by the Principle of Mathematical Induction, we can say that, P(n): n³ + 2n is divisible by 3 is true.

PMI in Computer Science

Mathematical induction is widely used in CS because many computer science concepts are naturally recursive or iterative, and induction mirrors that structure.

Correctness of Algorithms

• Proving loop invariants hold at every iteration.
• Showing recursive algorithms terminate and produce correct results.

Data Structures

• Proofs about trees, linked lists, and graphs often use structural induction (a variation of PMI).
• Example: “A binary tree with nnn nodes has n−1n-1n−1 edges.”

Complexity Analysis

• Proving formulas for time/space complexity, especially for recursive algorithms like Merge Sort.

Combinatorics in CS

• Counting problems, correctness of combinatorial formulas used in algorithms.

Formal Languages & Automata

• Proving properties of strings, grammars, and languages.

Related Articles

• Arithmetic Progression
• Geometric Progression

Solved Examples of Mathematical Induction

Example 1: For all n ≥ 1, prove that, 1² + 2² + 3²+....+n² = {n(n + 1) (2n + 1)} / 6

Solution:

Let the given statement be P(n),

For n = 1, the left side :
1 ² = 1

The right-hand side:
P(n):1^2+ 2^2 + 3^2+ \ldots+ n^2 = \frac{n(n + 1) (2n + 1)}{6} \\~\\ \text{For n=1} \\ P(1):\frac{1(1+1)(2×1+1)}{6} = 1        

Now, let's take a positive integer, k, and assume P(k) to be true i.e.,

1^2 + 2^2 + 3^2 +....+k^2 = \frac{k(k+1)(2k+1)}{6}        

We shall now prove that P(k + 1) is also true, so now we have,

P(k + 1) = P(k) + (k + 1)²  

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2
= (k+1) \frac{( 2k^2 + k) + 6(k+1)}{6}
=\frac{(k+1)(2k^2 +7k+6)}{6}
=\frac{(k+1) (k+2) (2k+3)}{6}
=\frac{(k+1) ((k+1)+1) (2(k+1) +1)}{6}

Thus P(k + 1) is true, whenever P(k) is true for all natural numbers. Hence, by the process of mathematical induction, the given result is true for all natural numbers.

Example 2: For all n ≥ 1, prove that, 1.2.3 + 2.3.4 + 3.4.5+...+n(n + 1) (n + 2) = {n (n + 1) (n + 2) ( n + 3)} / 4

Solution: 

Let the given statement be S(n),

S(n):1.2.3+ 2.3.4 + 3.4.5+\ldots+ n.(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n+3)}{4}\\ \ \\ \text{For n=1,} \\ S(1):\frac{1(1+1)(1+2)(1+3)}{4} = 6\\ \text{which is true.}

Now, let's take a positive integer, k, and assume S(k) to be true i.e.
S(k):1.2.3+ 2.3.4 + 3.4.5+\ldots+ k.(k+1)(k+2) = \frac{k(k+ 1)(k + 2)(k+3)}{4}

We shall now prove that  S(k + 1) is also true, so now we have,
S(k+1):S(k) + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)}{4} + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)+ \ 4(k+1)(k+2)(k+3)}{4} \\ \ \\ \Rightarrow S(k+1):  \frac{(k+1)(k+2)(k+3)(k+4)}{4}\\ \ \\ \Rightarrow S(k+1): \frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\}\{(k+1)+3\} }{4} \\ 

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 3: For all n ≥ 1, prove that, 1 + 3 + 5 +... + 2n - 1 = n²

Solution: 

Let the given statement be S(n), 
and S(n) = 1 + 3 + 5 +... + 2n - 1 = n²
For n = 1,
L.H.S = 2 (1) × 1 - 1 = 1
R.H.S = 1² = 1

Thus S(1) is true .

Now, let's take a positive integer, k, and assume S(k) to be true i.e.,
S(k) = 1+ 3 + 5+...+(2k - 1) = k² 

We shall now prove that  S(k + 1) is also true, so now we have,
1 + 3 + 5+...+ (2(k + 1) - 1) = (k + 1)² 

L.H.S = 1 + 3 + 5 + .... (2k - 1 ) + 2k + 2 - 1

⇒ L.H.S = S(k) + 2k + 1
⇒ L.H.S = k² + 2k + 1
⇒ L.H.S = (k + 1)² 
⇒ L.H.S = R.H.S

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 4: For all n ≥ 1, prove that, 1.2 + 2.3 + 3.4 +...+ n(n + 1) = {n(n + 1)(n + 2)} / 3

Solution: 

Let the given statement be S(n),
\ S(n):1.2+ 2.3 + 3.4+ ……+ n.(n+1) = \frac{n(n + 1)(n + 2)}{3}\\ \ \\ \text{for n=1,} \\ \ S(1) : \frac{1(1+1)(1+2)}{3} = 2\\ \text{which is true.}

Now, let's take a positive integer, k, and assume S(k) to be true i.e.,
S(k):1.2+ 2.3 + 3.4+ ……+ k.(k+1) = \frac{k(k+ 1)(k + 2)}{3} \ \\       

We shall now prove that S(k + 1) is also true, so now we have,
S(k+1) : S(k) + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) : \frac{k(k+ 1)(k + 2)}{3} + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) :\frac{k(k+ 1)(k + 2)+ 3(k+1)(k+2)}{3} \\ \ \\ \Rightarrow S(k+1) :\frac{(k+1)(k+2)(k+3)}{3}\\ \ \\ \Rightarrow S(k+1) :\frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\} }{3}        

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 5: Prove that a_n = a₁ + (n - 1) d is the general term of any arithmetic sequence.

Solution: 

For n = 1,
a_n = a₁ + (1 - 1) d = a₁
so the formula holds true for n = 1

Let us assume that the formula a_k = a₁ + (k - 1) is true for all natural numbers. 

We shall now prove that the formula is also true for k+1, so now we have,
a_{k + 1} = a₁ + [(k + 1) - 1] d = a₁ + k · d.

We assumed that a_k = a₁ + (k - 1) d, and by the definition of an arithmetic sequence a_{k+ 1} - a_k = d,

Then, a_{k + 1} - a_k 
= (a₁ + k · d) - (a1 + (k - 1)d)
= a₁ - a₁ + kd - kd + d
= d

Thus the formula is true for k + 1, whenever it is true for k. And we initially showed that the formula is true for n = 1. Thus the formula is true for all natural numbers.

Practice Questions on the Principle of Mathematical Induction

Question 1: Prove that for all natural numbers n, the following holds: 1 + 2 + 3 + . . . + n = [n(n + 1)]/2.

Question 2: Show that for all integers n ≥ 1 : 1³ + 2³ + 3³ + . . . + n³ = [{n(n + 1)}/2]².

Question 3: Verify that for all natural numbers n: 2ⁿ > n².

Question 4: Prove that for all integers n ≥ 1: 1×2 + 2×3 + 3×4 + ⋯ + n(n+1) = [n(n+1)(n+2)]/3.

Question 5: Show that for all integers n ≥ 1: 7ⁿ − 4ⁿ is divisible by 3.