- Category: Mathematics

**Mathematical Induction** is a technique of proving a statement, theorem or formula which is thought to be true, for each and every natural number n. By generalizing this in form of a principle which we would use to prove any mathematical statement is ‘**Principle of Mathematical Induction**‘.

For example: 1^{3} +2^{3} + 3^{3} + ….. +n^{3} = (n(n+1) / 2)^{2}, the statement is considered here as true for all the values of natural numbers.

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:

**Step 1: ** Check whether the given statement is true for n = 1.

**Step 2: **Assume that given statement P(n) is also true for n = k, where k is any positive integer.

**Step 3: ** Prove that the result is true for P(k+1) for any positive integer k.

If the above-mentioned conditions are satisfied, then it can be concluded that P(n) is true for all n natural numbers.

The first step of the principle is a factual statement and the second step is a conditional one. According to this if the given statement is true for some positive integer k only then it can be concluded that the statement P(n) is valid for n = k + 1.

This is also known as the inductive step and the assumption that P(n) is true for n=k is known as the inductive hypothesis.

**Example 1: ****Prove that the sum of cubes of n natural numbers is equal to ( n(n+1) ^{2})^{2} for all n natural numbers.**

**Solution**:

In the given statement we are asked to prove:

1^{3}+2^{3}+3^{3}+⋯+n^{3 }= (n(n+1)^{2})^{2}

Step 1: Now with the help of the principle of induction in math let us check the validity of the given statement P(n) for n=1.

P(1)=(1(1+1)^{2})^{2} = 1 This is true.

Step 2: Now as the given statement is true for n=1 we shall move forward and try proving this for n=k, i.e.,

1^{3}+2^{3}+3^{3}+⋯+k^{3}= (k(k+1)^{2})^{2} .

Step 3: Let us now try to establish that P(k+1) is also true.

1^{3}+2^{3}+3^{3}+⋯+k^{3}+(k+1)^{3} = (k(k+1)^{2})^{2}+(k+1)^{3}

⇒1^{3}+2^{3}+3^{3}+⋯+k^{3}+(k+1)^{3}=k^{2}(k+1)^{4}+(k+1)^{3}

= k^{2}(k+1)^{2}+4((k+1)^{3})^{4}

=(k+1)^{2}(k^{2}+4(k+1))^{4}

=(k+1)^{2}(k^{2}+4k+4)^{4}

= (k+1)^{2}((k+2)^{2})^{4}

=(k+1)^{2}(k+1+1)^{2})^{4}

=(k+1)^{2}((k+1)+1)^{2})^{4}

**Example 2**: **Show that 1 + 3 + 5 + … + (2n−1) = n ^{2}**

**Solution**:

**Step 1: **Result is true for n = 1

That is 1 = (1)^{2} (True)

**Step 2**: Assume that result is true for n = k

1 + 3 + 5 + … + (2k−1) = k^{2}

**Step 3**: Check for n = k + 1

i.e. 1 + 3 + 5 + … + (2(k+1)−1) = (k+1)^{2}

We can write the above equation as,

1 + 3 + 5 + … + (2k−1) + (2(k+1)−1) = (k+1)^{2}

Using step 2 result, we get

k^{2} + (2(k+1)−1) = (k+1)^{2}

k^{2} + 2k + 2 −1 = (k+1)^{2}

k^{2} + 2k + 1 = (k+1)^{2}

(k+1)^{2} = (k+1)^{2}

L.H.S. and R.H.S. are same.

So the result is true for n = k+1

By mathematical induction, the statement is true.

We see that the given statement is also true for n=k+1. Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n.