The procedure of proof should follow the following format. The answer should lik
ID: 3732601 • Letter: T
Question
The procedure of proof should follow the following format. The answer should like that.
Example
Consider the sequence a1, a2, a3,... given by a1= 2, a2= 5, and ai= ai-1+ 2ai-2for all i3.
Prove that, for all i1:ai= (1/3)·(7·2i-1+ (-1)i)
Proof: We prove the statement by induction on i.
Base Case:
We prove that ai= (1/3)·(7·2i-1+ (-1)i) when i=1 and i=2.
When i=1: we are given that a1= 2,and we see that (1/3)·(7·2i-1+ (-1)i) = (1/3)·(7-1) = 2, as required.
When i=2: we are given that a2= 5,and we see that (1/3)·(7·2i-1+ (-1)i) = (1/3)·(7·2+1) = (1/3)·15 = 5, as require
Inductive step: We prove that, for all k2,if aj= (1/3)·(7·2j-1+ (-1)j) for all j{1,...,k},then ak+1= (1/3)·(7·2k+ (-1)k+1).
(1)Let k be an arbitrary positive integer such that k2.Assume thataj= (1/3)·(7·2j-1+ (-1)j) for all j{1,...,k}.
(2)Since k2, we know that k+13, so the given recursive formula states that ak+1= ak+ 2ak-1.
(3)From (1), we know that ak= (1/3)·(7·2k-1+ (-1)k) and ak-1= (1/3)·(7·2k-2+ (-1)k-1).
(4)From (2) and (3), it follows that ak+1= (1/3)·(7·2k-1+ (-1)k) + 2·(1/3)·(7·2k-2+ (-1)k-1)= (1/3)·(7·2k-1+ (-1)k) +(1/3)·(7·2k-1+ 2·(-1)k-1)= (1/3)·(7·2k-1+ (-1)k) +(1/3)·(7·2k-1-2·(-1)k)= (1/3)·(7·2k-1+ 7·2k-1+ (-1)k- 2·(-1)k)= (1/3)·(7·2k- (-1)k) = (1/3)·(7·2k+ (-1)k+1) , which completes the inductive step.
4. A sequence of integers '12,13, is defined by -3,T2-7 and ln-trn-i-urn-2 lor n > 3. 6xn-2 for n >3 Prove that in 2n + 3n-i is an explicit formula for the sequence for all n > 1.Explanation / Answer
Proof: We proove the statement by induction on "n"
BaseCase:We prove thet xn = 2n+3n-1 when i=1,i=2.
Case 1:When n=1 : We are given that x1 = 3 and we see that 2n+3n-1 = 21+31-1 =2+30 =2+1 = 3, as required
Case 2: When n=2 : We are given that x2 = 7 and we see that 2n+3n-1 = 22+32-1 =4+31 =4+3 = 7, as required
Induction Step: We prove that, for all k>=2,if xm = 2m+3m-1 for all m{1,...,k}, then xk+1 = 2k +3k-1+1
1. Let k be an arbitary positive integer such that k>=2. Assume that xm = 2m+3m-1 for all m{1,...,k}.
2. Since k>=2, we know that k+1 > = 3, so the given recursive firmula states that xk+1 = 5xk-1-6xk-2+1
3. From (1) , we know that xk = 2k+3k-1 and xk-1 =2k-1+3k-2.
4. From (2) and (3), it follows that xk+1 = 2k +3k-1 +1, whichcompletes the inductive step.
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