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Geven the recmence nelation = 10r,-25 al= L, = 10. We meed to prove that , =5"(1+R) for all21. To do so, the inductive step shold be Xn = 5"(1+ n) and show that x+1 = 5"+'(2 +n). By first principle of mathematical induction, we need to suppose the above X*+1 = 10x - 25x-1 for k

Geven the recmence nelation = 10r,-25 al= L, = 10. We meed to prove that , =5"(1+R) for all21. To do so, the inductive step shold be Xn = 5"(1+ n) and show that x+1 = 5"+'(2 +n). By first principle of mathematical induction, we need to suppose the above X*+1 = 10x - 25x-1 for k

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Given the recurence relation x = 10r, - 25x-1 and xg = 1,x = 10. We need to prove that x, = 5"(1 + n) for all n 2 1. To do so, the inductive step should be Xn = 5"(1+n) and show that xm+1 = 5"+1(2+n). By first principle of mathematical O induction, we need to suppose the above Xg+1 = 10xx – 25xg -1 for k < n and show that x+1 = 10xn – 25xn-1 · By Strong induction, we need to suppose the above None of these Xg = 5*(1+ k) for k < n and show that x, = 5"(1+n). By Strong induction, we need to suppose the above Xn = 5"(1+) and show that xn+1 = 5"+1(2 +n). By Strong induction, we need to suppose the above
Transcribed Image Text:Given the recurence relation x = 10r, - 25x-1 and xg = 1,x = 10. We need to prove that x, = 5"(1 + n) for all n 2 1. To do so, the inductive step should be Xn = 5"(1+n) and show that xm+1 = 5"+1(2+n). By first principle of mathematical O induction, we need to suppose the above Xg+1 = 10xx – 25xg -1 for k < n and show that x+1 = 10xn – 25xn-1 · By Strong induction, we need to suppose the above None of these Xg = 5*(1+ k) for k < n and show that x, = 5"(1+n). By Strong induction, we need to suppose the above Xn = 5"(1+) and show that xn+1 = 5"+1(2 +n). By Strong induction, we need to suppose the above
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