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Anonymous Thu 09 Apr 15:55:03 2020 No.11548134 View Reply Original Report

Quoted By:

Can someone help me solve this question please? I got the basis step done, as well as the induction hypothesis assume p(k) which is bk = 11 is true. Now for the induction step, I'm kind of stuck. b(k+1) = 11, and I don't know where to go from here.

Anonymous

Anonymous Thu 09 Apr 2020 16:57:20 No.11548373 Report

Quoted By: >>11548516

base: b_3 = (11 + 11 + 11) / 3 = 11
I.H.: b_k = 11 for all n <= k
I.S.: b_{k+1} = (b_k + b_{k-1} + b_{k-2})/3
= (11 + 11 + 11)/3
= 11
Thus, the I.S. holds and the proof is shown.

Anonymous

Anonymous Thu 09 Apr 2020 17:40:37 No.11548516 Report

Quoted By: >>11548521

>>11548373
Ah, I didn't realize we could just say b_k = 11 for all n <= k.

Thanks for the help, it's much appreciated.

Anonymous

Anonymous Thu 09 Apr 2020 17:42:27 No.11548521 Report

Quoted By: >>11548610

>>11548516
Google Strong Induction

Anonymous

Anonymous Thu 09 Apr 2020 18:21:43 No.11548610 Report

Quoted By:

>>11548521
So if I understand this correctly, to use strong induction with say a amount of previous cases that you need to be true, you need to show at least a cases in your basis right? And then in your inductive hypothesis you write "Assume P(k) is true for all n , k-a <= n <= k"? and then go on with your inductive step right?

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