Example-1 :-  For all n ≥ 1, prove that :

Example-2 :-  Prove that 2ⁿ > n for all positive integers n.

  Let P(n):  2n > n When n = 1, 21 >1. 
  Hence P(1) is true. 

  Assume that P(k) is true for any positive integer k, i.e., 2k  > k ... (1) 

  We shall now prove that P(k+1) is true whenever P(k) is true. 
  Multiplying both sides of (1) by 2, 
  we get 2. 2k > 2k i.e., 2k+1 > 2k = k + k > k + 1 

  Therefore, P(k + 1) is true when P(k) is true. 
  Hence, by principle of mathematical induction, P(n) is true for every positive integer n. 
   

Example-3 :-  For all n ≥ 1, prove that :

Example-4 :-  For every positive integer n, prove that 7ⁿ – 3ⁿ is divisible by 4.

  We can write P(n) :  7n – 3n is divisible by 4. 
  We note that P(1): 71 – 31 = 4 which is divisible by 4. 
  Thus P(n) is true for n = 1 
  Let P(k) be true for some natural number k, i.e., P(k) : 7k – 3k  is divisible by 4. 
  We can write 7k – 3k = 4d, where d ∈ N. 

  Now, we wish to prove that P(k + 1) is true whenever P(k) is true. 
  Now 7k+1 – 3k+1 
= 7k+1 – 7.3k + 7.3k – 3k+1 
= 7(7k – 3k) + (7 – 3)3k  
= 7(4d) + (7 – 3)3k 
= 7(4d) + 4.3k  
= 4(7d + 3k) 
        
  From the last line, we see that 7k+1 – 3k+1 is divisible by 4. 
  Thus, P(k + 1) is true when P(k) is true. 
  Therefore, by principle of mathematical induction the statement is true for every positive integer n. 
   

Example-5 :-  Prove that : (1 + x)ⁿ ≥ (1 + nx), for all natural number n, where x > – 1.

  Let P(n) be the given statement, i.e., P(n): (1 + x)n ≥ (1 + nx), for x > – 1. 
  We note that P(n) is true when n = 1, since ( 1+x) ≥  (1 + x) for x > –1 

  Assume that P(k): (1 + x)k ≥ (1 + kx), x > – 1 is true. ... (1) 

  We want to prove that P(k + 1) is true for x > –1 whenever P(k) is true. ... (2) 
  Consider the identity (1 + x)k+1 = (1 + x)k (1 + x) 
  Given that x > –1, so (1+x) > 0. 
  Therefore , by using (1 + x)k ≥ (1 + kx), 
  we have (1 + x)k+1 ≥ (1 + kx)(1 + x) 
  i.e. (1 + x)k+1 ≥  (1 + x + kx + kx2).    ... (3)
  Here k is a natural number and x2 ≥ 0 so that kx2 ≥ 0. 
  Therefore (1 + x + kx + kxsup>2) ≥ (1 + x + kx), and 
  so we obtain (1 + x)k+1 ≥ (1 + x + kx) 
  i.e. (1 + x)k+1 ≥  [1 + (1 + k)x] 

  Thus, the statement in (2) is established. 
  Hence, by the principle of mathematical induction, P(n) is true for all natural numbers.
   

Example-6 :-  Prove that : 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24, for all n ∈ N.

  Let the statement P(n) be defined as P(n) : 2.7n + 3.5n – 5  is divisible by 24. 
  We note that P(n) is true for n = 1, since 2.7 + 3.5 – 5 = 24, which is divisible by 24. 

  Assume that P(k) is true i.e. 2.7k + 3.5k – 5 = 24q, when q ∈ N ... (1) 

  Now, we wish to prove that P(k + 1) is true whenever P(k) is true. 
  We have 2.7k+1 + 3.5k+1 – 5 
= 2.7k . 71 + 3.5k . 51 – 5 
= 7 [2.7k + 3.5k – 5 – 3.5k + 5] + 3.5k . 5 – 5 
= 7 [24q – 3.5k + 5] + 15.5k –5 
= 7 × 24q – 21.5k + 35 + 15.5k – 5 
= 7 × 24q – 6.5k + 30 
= 7 × 24q – 6 (5k – 5) 
= 7 × 24q – 6 (4p) [(5k – 5) is a multiple of 4] 
= 7 × 24q – 24p 
= 24 (7q – p) 
= 24 × r; 
  r = 7q – p, is some natural number. ... (2) 

  The expression on the R.H.S. of (1) is divisible by 24. 
  Thus P(k + 1) is true whenever P(k) is true. 
  Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.
    

Example-7 :-  Prove that : 1² + 2² + ... + n² > n³/3, n ∈ N

  Let P(n) be the given statement.
  i.e., 12 + 22 + ... + n2 > n3/3, n ∈ N
  We note that P(n) is true for n = 1 since 12 > 13/3

  Assume that P(k) is true 
  i.e. P(k) : 12 + 22 + ... + k2  > k3/3 ...(1)

  We shall now prove that P(k + 1) is true whenever P(k) is true. 
  We have 12 + 22 + 32 + ... + k2 + (k + 1)2 
= (12 + 22 + 32 + ... + k2) + (k + 1)2 > k3/3 + (k + 1)2  [by (1)]
= 1/3 [k3 + 3k2 + 6k + 3]
= 1/3 [(k + 1)3 + 3k + 2] > 1/3 (k+1)3

  Therefore, P(k + 1) is also true whenever P(k) is true.
  Hence, by mathematical induction P(n) is true for all n ∈ N. 
   

Example-8 :-  Prove the rule of exponents (ab)ⁿ = aⁿbⁿ by using principle of mathematical induction for every natural number.

  Let P(n) be the given statement i.e. P(n) : (ab)n = anbn. 
  We note that P(n) is true for n = 1 
  since (ab)1 = a1b1.
        
  Let P(k) be true, i.e., (ab)k = akbk ... (1) 
  We shall now prove that P(k + 1) is true whenever P(k) is true. 
   
  Now, we have (ab)k+1
= (ab)k (ab)
= (ak bk) (ab)              [by (1)] 
= (ak . a1) (bk . b1)
= ak+1 . bk+1 
  
  Therefore, P(k + 1) is also true whenever P(k) is true. 
  Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.