Proof by mathematical induction

Algebra

Proof by mathematical induction

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 $$2^{2n}$$ +6n+2 is dividable by 6.
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Re: Proof by mathematical induction

nevermind I solved it
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Re: Proof by mathematical induction

For those that are interested you can use "proof by induction" to prove that "$$2^{2n}+ 6n+ 2$$ 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 $$2^2+ 6+ 2= 4+ 6+ 2= 12= 2(6)$$.

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 $$2^{2(k+1)}+ 6(k+1)+ 2= 2^{2k+ 2}+ 6k+ 6+ 2= (4(2^{2k})+ 6k+ 2)+ 6= (2^{2k}+ 6k+ 2)+ 3(2^k)+ 6$$. Since we are assuming $$2^{2k}+ 6k+ 2$$ is divisible by 6, that is equal to 6m for some integer m, since k is greater than or equal to 1, $$3(2^k)$$ is divisible by 6.
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