help with proof

4chan, i require your assistance:
prove that 2^n+(-1)^(n+1) for n>=2 is always divisible by 3.

Any pointers on where to start? I know your not going to do my homework for me, this isn't actually the homework question, but if i can prove this then it proves what im trying to prove. Just some tips maybe ? thanks.

Just say "It's in the bible, so it must be true"

every member in the sequence 2^n+(-1)^(n+1) for n>=2
is 3 times every member in this reccurance relation:
a_n=a_(n-1)+2*a_(n-2)

now where to from here...

Proof by induction:

2^n+(-1)^(n+1) - (2^(n-1)+(-1)^(n))
= 2^n-2^(n-1) + (-1)^(n+1)-(-1)^n
= 2^(n-1) + (-1)^n * (-1-1)
= 2^(n-1) - 2 * (-1)^n
= 2^(n-1) + 2 * (-1)^(n-1)
= 2 * (2^(n-2) + (-1)^(n-1))
So...
2^n+(-1)^(n+1) = (2^(n-1)+(-1)^(n)) + 2 * (2^(n-2)+(-1)^(n-1))

Since the relation holds for n=2 and n=3 (verify manually), the proof is complete.

2 is congruent to -1 modulo 3, so 2^n is congruent to (-1)^n modulo 3. (-1)^n + (-1)^(n+1) = (-1)^n(1 + -1) = 0, so 2^n + (-1)^(n+1) is congruent to 0 modulo 3.

>>4
>>5
Thanks guys, i think i got now :)

yeah, 4chan has some good jews

2^n + (-1)^(n+1) (mod 3) = 2^n + 2^(n+1) (mod 3) = 2^n(1 + 2) (mod 3) = 2^n(3) (mod 3) = 0 (mod 3)

>>7
wrong... proof: the only good jew is a dead jew

Don't call me gay, but I need some mary jay!

Marijuana MUST be legalized.