For the left hand side of p(1) they are not looking for number but they are looking for something in term of f so what is the answer? Should it be f1? f = 2 + 13 for every integer n 2 1(by mathematical induction):se f,, f2, f3 is a sequence that satisfies the recurrence relationf, = f, -1 + 2 for each integer k 2 2, with initial condition f, = 1.eed to show that when the sequence f,, f,, f3is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above.t the property P(n) be the equation f = 2" + - 3. We will show that P(n) is true for every integer n 2 1.- that P(1) is true:eft-hand side of P(1) is 2- 3, which equals 1The right-hand side of P(1) is 1Since the left-hand and right-v that for each integer k 2 1, if P(k) is true, then P(k + 1) is true:R be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = 9+13where f,, f2, f3... is a sequence defined by the ref, = f, -1 + 2* for each integer k 2 2, with initial condition f, = 1.s is the inductive hypothesis.]must show that P(k + 1) is true. In other words, we must show that f,k +2Now the left-hand side of P(k + 1) isk +1 = f, + 2k +1by definition of f, fa, f.gk +1- 3+ 2k + 1by the laws of algebra= 2 -っk+1- 3- 3gk+2-3%3Dby the laws of algebrad this is the right-hand side of P(k + 1). Hence the inductive step is complete.Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is comniete 1

Answered: Image /qna-images/answer/93edfb94-2f7f-4fed-862b-4184bb77093a.jpg

y mathematical induction): f, f2, f3,... is a sequence that satisfies the recurrence relation f, = f, - + 2 for each integer k 2 2, with initial condition f, = 1. d to show that when the sequence f,, f,, fa,... is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above. he property P(n) be the equation f = 2" + - - 3. We will show that P(n) is true for every integer n 2 1. hat P(1) is true: t-hand side of P(1) is 2° -3 which equals 1 The right-hand side of P(1) is 1 Since the left-hand and right- hat for each integer k 2 1, if P(k) is true, then P(k + 1) is true: e any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = |クk+1-3 where f,, f,, fa,... is a sequence defined by the r fy = f -1+ 2* for each integer k 2 2, with initial condition f, = 1. s the inductive hypothesis.] ust show that P(k + 1) is true. In other words, we must show that f, *+2 -3 + 1 Now the left-hand side of P(k + 1) is k +1 = f, + 2k +1 by definition of f., fa, f, .v 2k +1 - 3 + 2k + 1 %3D by the laws of algebra 2k+1-3 = 2. - 3 2k+2-3 by the laws of algebra this is the right-hand side of P(k + 1). Hence the inductive step is complete. "s, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete 1

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.6: Applications And The Perron-frobenius Theorem

Problem 57EQ