Proof by mathematical induction

Algebra

Proof by mathematical induction

Postby Guest » Mon Apr 27, 2020 2:15 pm

I don't know if this is the right section but here we go.

Prove (using mathematical induction) that for every natural number n, it's true that [tex]2^{2n}[/tex] +6n+2 is dividable by 6.
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Re: Proof by mathematical induction

Postby Guest » Mon Apr 27, 2020 3:00 pm

nevermind I solved it :roll:
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Re: Proof by mathematical induction

Postby Guest » Mon May 04, 2020 1:36 pm

For those that are interested you can use "proof by induction" to prove that "[tex]2^{2n}+ 6n+ 2[/tex] is divisible by 6" for every natural number n as follows.

First prove that it is true for n= 1. When n= 1 this is [tex]2^2+ 6+ 2= 4+ 6+ 2= 12= 2(6)[/tex].

Now prove that if it is true for any n= k then it is also true for n= k+ 1. When n= 1 this is [tex]2^{2(k+1)}+ 6(k+1)+ 2= 2^{2k+ 2}+ 6k+ 6+ 2= (4(2^{2k})+ 6k+ 2)+ 6= (2^{2k}+ 6k+ 2)+ 3(2^k)+ 6[/tex]. Since we are assuming [tex]2^{2k}+ 6k+ 2[/tex] is divisible by 6, that is equal to 6m for some integer m, since k is greater than or equal to 1, [tex]3(2^k)[/tex] is divisible by 6.
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