Principle of Mathematical Induction with Examples

The concept of mathematical induction is the art of proving a statement, theorem or a 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’.

Principle of Mathematical Induction

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:

i) Check whether the given statement is true for n=1.

ii) If P(1) is true, then check if the given statement is true for n=k, then we can say that the given statement P(n) is also true for n=k+1.We can say thatP(k+1) is true if P(k) is true 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.

This is the Principle of Mathematical Induction.

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.

Moving on to mathematical Induction examples let us have a look at certain statements and let us try to prove them using the concept of mathematical induction.

Example

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

Solution: In the given statement we are asked to prove:

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

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

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

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

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

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

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

\(~~~~~~~~~~~~~~~~~\)\(1^3+2^3+3^3+⋯+ k^3+(k+1)^3\)=\(\frac{k^2 (k+1)^2}{4}+(k+1)^3\)

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

\(~~~~~~~~~~~~~~~~~\)=\( \frac {(k+1)^2 (k^2+4(k+1))}{4}\)

\(~~~~~~~~~~~~~~~~~\)=\( \frac{(k+1)^2(k^2 + 4k+4)}{4}\)

\(~~~~~~~~~~~~~~~~~\)= \( \frac{(k+1)^2(k+2)^2}{4}\)

\(~~~~~~~~~~~~~~~~~\)=\( \frac{(k+1)^2(k+1+1)^2}{4}\)

\(~~~~~~~~~~~~~~~~~\)=\( \frac{(k+1)^2((k+1)+1)^2}{4}\)

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 number sn.

Hence we have seen how easy it becomes to prove any statement using this principle. To know more about math visit us at www.byjus.com. We wish you a Happy learning!


Practise This Question

10(2n1)+1 is divisible by