We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities
Multiply both sides of the above inequality by 3 Let n = 1 and calculate 3 1 and 1 2 and compare themģ is greater than 1 and hence p (1) is true. STEP 1: We first show that p (1) is true. Prove that 3 n > n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n > n 2 for n a positive integer greater than 2. Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true. We now consider the algebraic expression (k + 1) 3 + 2 (k + 1) expand it and group like terms K 3 + 2 k = 3 M, where M is a positive integer. Prove that for any positive integer number n, n 3 + 2 n is divisible by 3 We have started from the statement P(k) and have shown thatġ 2 + 2 2 + 3 2 +. Set common denominator and factor k + 1 on the right side + k 2 = k (k + 1) (2k + 1)/ 6Īnd show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statementġ 2 + 2 2 + 3 2 +. STEP 1: We first show that p (1) is true.īoth sides of the statement are equal hence p (1) is true.Īnd show that p (k + 1) is true by adding k + 1 to both sides of the above statementġ + 2 + 3 +. Step 2: We assume that P (k) is true and establish that P (k+1) is also true Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.
Let us denote the proposition in question by P (n), where n is a positive integer. The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality…) is true for all positive integer numbers greater than or equal to some integer N. Several problems with detailed solutions on mathematical induction are presented.
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