Induction proof on a sequence A sequence a 0 </msub> , a 1 </

daktielti 2022-07-01 Answered
Induction proof on a sequence
A sequence a 0 , a 1 , is defined by letting a 0 = a 1 = 1 and a k = a k 1 + 2 a k 2 for all integers k > 1. Prove that for all integers k 0, a k = 2 k + 1 + ( 1 ) k 3 .
Base case: S(2)
Let k = 2. Then a 2 = 2 2 + 1 + ( 1 ) 2 3 = 9 3 = 3.
S(3)
Let k = 3. Then a 3 = 2 3 + 1 + ( 1 ) 3 3 = 15 3 = 5
Base is true.
Induction step: Assume S(k) and S ( k 1 ) are true. Then
S ( k ) = a k = 2 k + 1 + ( 1 ) k 3
and S ( k 1 ) = a k + 1 = 2 k + ( 1 ) k 1 3
Since S ( k ) = a k and S ( k 1 ) = a k 1 , by definition S ( k + 1 ) = S ( k ) + 2 S ( k 1 ).
2 k + 2 + ( 1 ) k + 1 3 = 2 k + 1 + ( 1 ) k 3 + 2 2 k + ( 1 ) k 1 3
Solving this should lead to that
k = 2
Therefore S ( k + 1 ) is true.
I think I've got the right idea but I'm not completely sure that I've done this correctly. Any help or suggestions are much appreciated!
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Answers (1)

Freddy Doyle
Answered 2022-07-02 Author has 20 answers
Step 1
S(k) must have a truth value. In this case it's the statement a k = 2 k + 1 + ( 1 ) k 3 . So you shouldn't write S ( k + 1 ) = S ( k ) + 2 S ( k 1 ) because that's adding statements!
Step 2
The question asks to show S(k) for k 0. So your base case must include k = 0. (Note that here I've abbreviated "S(k) is true" to "S(k)" – in the same way that I can abbreivate "it's true that it's raining" to "it's raining". But feel free to say "S(k) is true" for clarity.)
Step 3
The aim for the induction part is to (as I think you know) assume S ( k 1 ) and S(k), and then show S ( k + 1 ). The aim is then to do something like
a k + 1 = a k + 2 a k 1 (by definition) = 2 k + 1 + ( 1 ) k 3 + 2 k + ( 1 ) k 1 3 (by inductive hypothesis) = (some steps for you to work out) = 2 k + 2 + ( 1 ) ( k + 1 ) 3
i.e. S ( k + 1 ). In your proof you seemed to begin with assuming S ( k + 1 ), which is incorrect.

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