Please show work 5. For n,k € Z such that 0 ≤ k ≤n, define(3)Note that, by convention, 0!= 1.(a) Prove that () = 1, () = 1, andn!k!(n-k)!'(3) − (¹7¹) + (−¹)k-if(b) Use part (a) and induction to prove that (2) is a positive integer for all n, k Zsuch that 0≤k≤n.(c) Let x, y R. Prove that for every integer n ≥ 0,711≤k

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5. For n,k € Z such that 0 ≤ k ≤n, define (3). Note that, by convention, 0! = 1. (a) Prove that () = 1, () = 1, and n! k!(n - k)! n- (1)-(²7)+(-)) * if = 1≤k≤n-1. (b) Use part (a) and induction to prove that () is a positive integer for all n, k Z such that 0 ≤k≤n. (c) Let z, y € R. Prove that for every integer n 20, (x + y)² = Σ (1) ²¹-²², k=0

5. For n,k € Z such that 0 ≤ k ≤n, define (3). Note that, by convention, 0! = 1. (a) Prove that () = 1, () = 1, and n! k!(n - k)! n- (1)-(²7)+(-)) * if = 1≤k≤n-1. (b) Use part (a) and induction to prove that () is a positive integer for all n, k Z such that 0 ≤k≤n. (c) Let z, y € R. Prove that for every integer n 20, (x + y)² = Σ (1) ²¹-²², k=0

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.2: Properties Of Division

Problem 51E

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