At n=0 that looks like 2(6^0)-5(6^0)-2 = -5, a counterexample.
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It works for n=1 and n=2 so let's try to prove it for natural n>0.
I've been warned off using "mod" in answers so we'll kind of handwave around it.
We substitute n=m+1 so m is a non-negative integer, m >= 0.
[tex] 2(6^{2n})-5(6^n)-2= 2(6^{2m + 2}) - 5(6^{m+1}) -2 = 2(6^2)6^{2m} - 5(6)6^m - 2 = 72( 6^{2m} ) - 30 (6^m) - 2[/tex]
The term times 30 will always be a multiple of 10, so we need to check the rest of it is as well. The factor of 72=70+2 and the 70 part will always give a multiple of 10.
Also [tex]6^{2m}= (6^2)^m= 36^m [/tex] and that will have the same remainder as [tex]6^m[/tex] when divided by 10. In fact every power of 6 except the zeroth ends in 6, because 6 times 6 is 36.
So we must show
[tex]2( 6)- 2[/tex]
is a multiple of 10 which of course it is. We also need to check m=0, 2(1)-2=0, ok as well. That ends the proof.
