;;;; BEG of file "b2.soln.txt" ;;;; Dear Numbers&Polys Enthusiast-Scholars Agree to use braces for grouping (not indicating sets), e.g, b_{n+1} means "b sub n+1". I'll write this solution in more detail than you needed to write on the exam. PLEASE GRADE my writing, and let me know (via email) of improvements/corrections that you see. Bear in mind that I am limited by the format of simple email. Best Wishes, "Prof. Jonathan" ================================================================ B2: Define a sequence b_0,b_1,b_2, ... of integers 1: b_0 := 0 and b_1 := 3 and 2: b_{n+2} := 7·b_{n+1} - 10·b_n for n=0,1, ... . Use strong induction to prove, for all k >= 0, that 3(k): b_k = 5^k - 2^k . /------------------------------------------------------------\ / \ Proof: For k=0,1,... define R(k) := 5^k - 2^k . Base Cases: Observe that b_0 = 0 , by definition, = 1 - 1 = 5^0 - 2^0 = R(0) , again by defn. Similarly b_1 = 3 , by defn, = 5^1 - 2^1 = R(1) , by defn. ================================================================ Induction case: Fixing a natnum n, our goal is to establish that b_{n+2} equals R(n+2) using that (3(k)) holds for all k=0,1,2,...,n+1. Here goes! By defn, b_{n+2} = 7·b_{n+1} - 10·b_n . Courtesy equalities (3(n+1)) and (3(n)), 4: b_{n+2} = 7·[5^{n+1} - 2^{n+1}] - 10·[5^n - 2^n] . [That was the "strong induction" step.] Regrouping terms produces 7·[5^{n+1} - 2^{n+1}] = 7·5 · 5^n - 7·2 · 2^n and 10·[5^n - 2^n] = 10 · 5^n - 10 · 2^n . Gathering the powers of 5 in the righthand sides above thus yields [7·5 - 10] · 5^n = [35-10] · 5^n = 25 · 5^n . Similarly, gathering the powers of 2 above gives us [7·2 - 10] · 2^n = [14-10] · 2^n = 4 · 2^n . So these together with (4) say that 4': b_{n+2} = 5^2 · 5^n - 2^2 · 2^n Happily, this last equals R(n+2), which is a comfort. QED \ / \____________________________________________________________/ ;;;; END of file "b2.soln.txt" ;;;;