Example-1 :- For all n ≥ 1, prove that :
Example-2 :- Prove that 2n > n for all positive integers n.
Solution :-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.
Solution :-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.
Solution :-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.
Solution :-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
Solution :-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.
Solution :-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.