Important Questions: Principles of Mathematical Induction | Mathematics for Grade 11 PDF Download

Q1: Prove that 1 + 3 + 5 + … + (2n – 1) = n² using the principle of mathematical induction.
Ans: Given Statement: 1 + 3 + 5 + … + (2n – 1) = n²
Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n² , for n ∈ N
Note that P(1) is true, since
P(1) : 1 = 1²
Let P(k) is true for some k ∈ N,
It means that,
P(k) : 1 + 3 + 5 + … + (2k – 1) = k²
To prove that P(k + 1) is true, we have
1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k² + (2k + 1)
= k² + 2k + 1
By using the formula, the above form can be written as:
= (k + 1)2
Hence, P(k + 1) is true, whenever P(k) is true.
Therefore, P(n) is true for all n ∈ N is proved by the principle of mathematical induction.

Q2: Prove that 2ⁿ > n for all positive integers n by the Principle of Mathematical Induction
Ans: Assume that P(n): 2ⁿ > n
If n =1, 2¹>1. Hence P(1) is true
Let us assume that P(k) is true for any positive integer k,
It means that, i.e.,
2^k > k …(1)
We shall now prove that P(k +1) is true whenever P(k) is true.
Now, multiplying both sides of the equation (1) by 2, we get
2. 2^k > 2k
Now by using the property,
i.e., 2^k+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 is proved using the principle of mathematical induction.

Q3: Show that 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n by the Principle of Mathematical Induction.
Ans: Assume that P(n) be the given statement, that is
P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n.
It is noted that P (1) is true, since
P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1.
Let P(n) is true for some natural number k,
It means that
P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1
Inorder to prove P (k + 1) is true, we have
P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)!
= (k + 1)! – 1 + (k + 1)! × (k + 1)
Now, simplify the above form, we get
= (k + 1 + 1) (k + 1)! – 1 = (k + 2) (k + 1)! – 1 = ((k + 2)! – 1
Therefore, P (k + 1) is true, whenever P (k) is true.
Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Induction.