(3n 2) = (3n 2) = 1(3 1) 2 1 = (3k 2) =

Transcription

1 Question 1 Prove using mathematical induction that for all n 1, Solution (n ) = n(n 1) For any integer n 1, let P n be the statement that (n ) = n(n 1) Base Case The statement P 1 says that which is true 1 = 1( 1), Inductive Step Fix k 1, and suppose that P k holds, that is, (k ) = k(k 1) It remains to show that P k+1 holds, that is, ((k + 1) ) = (k + 1)((k + 1) 1) Therefore P k+1 holds ((k + 1) ) = ((k + 1) ) = (k + 1) = (k ) + (k + 1) k(k 1) = + (k + 1) k(k 1) + (k + 1) = = k k + 6k + ) = k + k + ) (k + 1)(k + ) = (k + 1)((k + 1) 1) = Thus, by the principle of mathematical induction, for all n 1, P n holds

2 Question Use the Principle of Mathematical Induction to verify that, for n any positive integer, 6 n 1 is divisible by Solution For any n 1, let P n be the statement that 6 n 1 is divisible by Base Case The statement P 1 says that = 6 1 = is divisible by, which is true Inductive Step Fix k 1, and suppose that P k holds, that is, 6 k 1 is divisible by It remains to show that P k+1 holds, that is, that 6 k+1 1 is divisible by 6 k+1 1 = 6(6 k ) 1 = 6(6 k 1) = 6(6 k 1) + By P k, the first term 6(6 k 1) is divisible by, the second term is clearly divisible by Therefore the left hand side is also divisible by Therefore P k+1 holds Thus by the principle of mathematical induction, for all n 1, P n holds

3 Question Verify that for all n 1, the sum of the squares of the first n positive integers is given by the formula (n) n(n + 1)(n + 1) = Solution For any integer n 1, let P n be the statement that (n) = n(n + 1)(n + 1) Base Case The statement P 1 says that which is true 1 + = (1)((1) + 1)((1) + 1) = () =, Inductive Step Fix k 1, and suppose that P k holds, that is, (k) = k(k + 1)(k + 1) It remains to show that P k+1 holds, that is, ((k + 1)) = (k + 1)((k + 1) + 1)((k + 1) + 1) ((k + 1)) = (k + ) On the other side of P k+1, = (k) + (k + 1) + (k + ) k(k + 1)(k + 1) = + (k + 1) + (k + ) (by P k ) k(k + 1)(k + 1) = + (k + 1) + (k + ) = k(k + 1)(k + 1) + (k + 1) + (k + ) = k(8k + 6k + 1) + (k + k + 1) + (k + 8k + ) = (8k + 6k + k) + (1k + 1k + ) + (1k + k + 1) = 8k + 0k + 7k + 1 (k + 1)((k + 1) + 1)((k + 1) + 1) = (k + 1)(k + + 1)(k + + 1)

4 Therefore P k+1 holds = (k + 1)(k + )(k + ) = (k + k + )(k + ) = 8k + 0k + 7k + 1 Thus, by the principle of mathematical induction, for all n 1, P n holds

5 Question Consider the sequence of real numbers defined by the relations x 1 = 1 and x n+1 = 1 + x n for n 1 Use the Principle of Mathematical Induction to show that x n < for all n 1 Solution For any n 1, let P n be the statement that x n < Base Case The statement P 1 says that x 1 = 1 <, which is true Inductive Step Fix k 1, and suppose that P k holds, that is, x k < It remains to show that P k+1 holds, that is, that x k+1 < x k+1 = 1 + x k < 1 + () = 9 = < Therefore P k+1 holds Thus by the principle of mathematical induction, for all n 1, P n holds

6 Question Show that n! > n for n 7 Solution For any n 7, let P n be the statement that n! > n Base Case The statement P 7 says that 7! = = 00 > 7 = 187, which is true Inductive Step Fix k 7, and suppose that P k holds, that is, k! > k It remains to show that P k+1 holds, that is, that (k + 1)! > k+1 (k + 1)! = (k + 1)k! > (k + 1) k (7 + 1) k = 8 k > k = k+1 Therefore P k+1 holds Thus by the principle of mathematical induction, for all n 1, P n holds

7 Question 6 Let p 0 = 1, p 1 = cos θ (for θ some fixed constant) and p n+1 = p 1 p n p n1 for n 1 Use an extended Principle of Mathematical Induction to prove that p n = cos(nθ) for n 0 Solution For any n 0, let P n be the statement that p n = cos(nθ) Base Cases The statement P 0 says that p 0 = 1 = cos(0θ) = 1, which is true The statement P 1 says that p 1 = cos θ = cos(1θ), which is true Inductive Step Fix k 0, and suppose that both P k and P k+1 hold, that is, p k = cos(kθ), and p k+1 = cos((k + 1)θ) It remains to show that P k+ holds, that is, that p k+ = cos((k + )θ) We have the following identities: cos(a + b) = cos a cos b sin a sin b cos(a b) = cos a cos b + sin a sin b Therefore, using the first identity when a = θ and b = (k + 1)θ, we have cos(θ + (k + 1)θ) = cos θ cos(k + 1)θ sin θ sin(k + 1)θ, and using the second identity when a = (k + 1)θ and b = θ, we have Therefore, cos((k + 1)θ θ) = cos(k + 1)θ cos θ + sin(k + 1)θ sin θ p k+ = p 1 p k+1 p k = (cos θ)(cos((k + 1)θ)) cos(kθ) Therefore P k+ holds = (cos θ)(cos((k + 1)θ)) + (cos θ)(cos((k + 1)θ)) cos(kθ) = cos(θ + (k + 1)θ) + sin θ sin(k + 1)θ + (cos θ)(cos((k + 1)θ)) cos(kθ) = cos((k + )θ) + sin θ sin(k + 1)θ + (cos θ)(cos((k + 1)θ)) cos(kθ) = cos((k + )θ) + sin θ sin(k + 1)θ + cos((k + 1)θ θ) sin(k + 1)θ sin θ cos(kθ) = cos((k + )θ) + cos(kθ) cos(kθ) = cos((k + )θ) Thus by the principle of mathematical induction, for all n 1, P n holds

8 Question 7 Consider the famous Fibonacci sequence {x n } n=1, defined by the relations x 1 = 1, x = 1, and x n = x n1 + x n for n (a) Compute x 0 (b) Use an extended Principle of Mathematical Induction in order to show that for n 1, [( ) x n = 1 n ( ) n ] (c) Use the result of part (b) to compute x Solution (a) 1, 1,,,, 8, 1, 1,,, 89, 1,, 77, 610, 987, 197, 8, 181, 676 (b) For any n 1, let P n be the statement that [( x n = Base Case The statement P 1 says that x 1 ) n ( 1 + ) 1 ( [ = [ = 1 [ = 1 = 1, which is true The statement P says that x ] ] 1 + ) ( 1 ) n ] 1 ) 1 ] 1 )

9 which is again true = 1 [( = 1 [( [ = 1 = 1, 1 + ) ( + 1 )] )] ] Inductive Step Fix k 1, and suppose that P k and P k+1 both hold, that is, x k 1 ) k, and x k ) k+1 ( 1 ) k+1 It remains to show that P k+ holds, that is, that x k+ 1 + ) k+ ( 1 ) k+ x k+ = x k + x k ) k ( + 1 ) k ( ) k+1 ( ) ( + ) ( 6 + ) ( 1 + ) k+1 ( 1 ) k ( 1 ) k+1 1 ) k+1 1 ) k ( ) 1 ) k ( 1 ) k ( ) 6 )

10 Therefore P k+ holds 1 + ) k+ ( 1 + ) ( ) ( 1 ) k+ 1 ) k ( 1 ) k ( 1 ) + 1 ) Thus by the principle of mathematical induction, for all n 1, P n holds (c) Plugging n = 0 in a calculator yields the answer quickly Induction Examples