Mathematical InductionThe process to establish the validity of an ordinary result involving natural numbers is the principle of mathematical induction. Working RuleLet n_{0} be a fixed integer. Suppose P (n) is a statement involving the natural number n and we wish to prove that P (n) is true for all n ≥n_{0}. 1. Basic of Induction: P (n_{0}) is true i.e. P (n) is true for n = n_{0}. 2. Induction Step: Assume that the P (k) is true for n = k. Example 1: Prove the follo2wing by Mathematical Induction: 1 + 3 + 5 +.... + 2n - 1 = n^{2}. Solution: let us assume that. P (n) = 1 + 3 + 5 +..... + 2n - 1 = n^{2}. For n = 1, P (1) = 1 = 1^{2} = 1 It is true for n = 1................ (i) Induction Step: For n = r, P (r) = 1 + 3 + 5 +..... +2r-1 = r^{2} is true......................... (ii) Adding 2r + 1 in both sides P (r + 1) = 1 + 3 + 5 +..... +2r-1 + 2r +1 = r^{2} + (2r + 1) = r^{2} + 2r +1 = (r+1)^{2}..................... (iii) As P(r) is true. Hence P (r+1) is also true. From (i), (ii) and (iii) we conclude that. 1 + 3 + 5 +..... + 2n - 1 =n^{2} is true for n = 1, 2, 3, 4, 5 ....Hence Proved. Example 2: Solution: For n = 1, It is true for n = 1. Induction Step: For n = r,................... (i) Adding (r+1)^{2} on both sides, we get As P (r) is true, hence P (r+1) is true. 1^{2} + 2^{2} + 3^{2} +......+ n^{2}= is true for n = 1, 2, 3, 4, 5 ..... Hence Proved. Example3: Show that for any integer n Solution: Let P (n) = 11^{n+2}+12^{2n+1} For n = 1, P (1) = 11^{3}+12^{3}=3059=133 x 23 So, 133 divide P (1).................. (i) Induction Step: For n = r, P (r) = 11^{r+2}+12^{2r+1}=133 x s............ (ii) Now, for n = r + 1, P (r+1) = 11^{r+2+1}+12^{2(r)+3}=11[133s-12^{2r+1}] + 144. 12^{2r+1} = 11 x 133s + 12^{2r+1}.133=133[11s+12^{2r+1}]=133 x t........... (iii) As (i), (ii), and (iii) all are true, hence P (n) is divisible by 133. Next TopicBinary Relation |