Principle of Mathematical Induction For Class 11 Notes are available for students here at BYJU’S for free. The notes of Class 11 Maths Principle of Mathematical Induction will help students to have a quick revision prior to their exams. These notes are provided with reference to NCERT textbooks and CBSE syllabus (2021-2022).
Deductive reasoning is one of the primary basis of mathematical thinking. Inductive reasoning is frequently used in mathematics and is a primary aspect of scientific reasoning, where collecting and analysing data is the norm. The word induction means generalisation from specific cases.
Principle of Mathematical Induction
There are disciplines in Maths (such as algebra), where the statements are generated in terms of n where n is a positive integer. Hence, such statements are proved by the principle of mathematical induction.
- Each of such statements are represented by P(n) which is linked with n (positive integer)
- Putting n = 1, we have to check the correctness of the statement
- Now, assuming P(k) as true, the truth value of P(k+1) is established, where k is a positive integer
For Example: P(n): 13 +23 + 33 + ….. +n3 = (n(n+1) / 2)2 is the true statement for the values of n, where n is a natural number.
Step 1: Put n = 1 and find the value of P(n). P(1) is true.
Step 2. Assume P(k) is true, for the positive integer, k.
Step 3: Now prove, P(k+1) is true whenever P(k) is true
List of Class 11 Maths Notes
Q.1: Prove that 2n > n for all positive integers n.
Solution: Let us take P(n): 2n > n
Now putting, n =1;
Which is true. So, P(1) is true
Let P(k) is true for any positive integer k,
2k > k … (1)
Now, we need to prove P(k+1) is true whenever P(k) is true.
Multiply by 2 on both sides on eq.(1)
2. 2k > 2k
2k + 1 > 2k = k + k > k + 1
Hence, P(k + 1) is true when P(k) is true.
Therefore, P(n) is true for every positive integer n, by principle of mathematical induction.
Q.2: Prove that 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2, by principle of mathematical induction.
Solution: Let us write the given statement as:
P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
Putting n = 1, we get;
P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2
The statement P(1) is true.
Let us assume that P(k) is true for a positive integer, k.
1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)
Now, we need to prove that P(k+1) is true, whenever P(k) is true.
1.2 + 2.22 + 3.22 + … + k.2k + (k+1)2k+1
= (k – 1) 2k + 1 + 2 + (k+1)2k+1 [By eq.(i)]
= 2k+1 [(k-1) + k+1)] + 2
= 2k+1 .2k + 2
= k2k+2 + 2
= [(k+1)-1]2k+2 + 2
Hence, P(k+1) is true whenever P(k) is true.
Thus, by the principle of mathematical induction, statement P (n) is true for all natural numbers (n).
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