by 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.