In this section 11th grade math, student will learn the new topic prove by mathematical induction.
In algebra or in other disciplines of mathematics, there are certain results or statements that are formulated in terms of 'n' is a positive integer. To prove such statements, well suited method based on the specific technique is known as the Principle of Mathematical Induction.

The Principle of Mathematical Induction

It states that :
i) P(1) is true.
ii) P(k) is true, then P(k+1) is also true, where k is a natural number.
Then the statement P(n) is true for every natural number n.

Examples on Prove by Mathematical Induction

Example 1 : Show that n^{3} + (n+1)^{3} + (n+2)^{3} is divisible by 9 for every natural number 'n'. Solution : P(n) : n^{3} + (n+1)^{3} +(n+2)^{3}
First we will verify P(1) is true.
For that , Put n=1 in P(n)
P(1) : 1^{3} + (1+1)^{3} + ( 1+2)^{3}
P(1) : 1 + 2^{3} + 3^{3}
P(1) : 1 + 8 + 27
P(1) : 36
According to the divisibility rule 36 is divisible by 9
∴ P(1) is true . ------(Equation 1)
Let us assume that,
P(k) is true. -------(equation 2)
that means P(k): k^{3} + (k+1)^{3}+(k+2)^{3} is divisible by 9.
Now we want to prove that P(k+1) is true.
So P(k+1) will be
P(k+1): (k+1)^{3} +(k+1+1)^{3} +(k+1+2)^{3}
P(k+1) : (k+1)^{3} +(k+2)^{3} +(k+3)^{3}
P(k+1): (k+1)^{3}+(k+2)^{3} + k^{3} + 9k^{2} +27k + 27
P(k+1):(k+1)^{3} +(k+2)^{3} + k^{3} + 9(k^{2}+3k +3)
From equation (2), (k+1)^{3}+ (k+2)^{3}+k^{3} is divisible by 9 and 9(k^{2}+ 3k +3)is also divisible by 9.
So, P(k+1) is also divisible by 9. -------(equation 3)
∴ From equation (1), (2) and (3) and by principle of mathematical induction,
n^{3} + (n+1)^{3}+ (n+2)^{3} is divisible by 9.

Example 2 : Show that the sum of the first 'n' odd natural numbers is n^{2}. Solution : Odd natural numbers are 1,3,5,... 2n-1
Sum = n^{2}
P(n): 1 + 3 + 5 + ...+(2n-1)=n^{2}
For n=1
P(1) = 2n-1<
= 2(1) -1
= 2 -1 =1
n^{2} = (1)^{2}
∴ P(1) is true. ----(equation 1)
Assume that P(k) is true.
So P(k): 1 + 3 + 5 +....(2k -1) = k^{2} is true. ----(equation 2)
Now we have to prove that P(k+1) is true.
P(k+1) : 1 + 3 + 5 +....+(2k-1) + [2(k+1)-1]
= k^{2} + (2k + 1)
= (k+1)^{2}
∴ P(k+1) is true whenever p(k) is true.
Hence by equation (1), (2) and by principle of mathematical induction P(n) is true for all natural number 'n'.