Question 1137651
Base case:  n=1 

{{{ 2^1 + (-1)^(2) = 2+1 = 3 }}}<br>

Hypothesis:  Assume  {{{2^n + (-1)^(n+1) }}} is divisible by 3 for n=k<br>

Step case:  Let n=k+1:

     {{{ V = 2^(k+1) + (-1)^((k+1)+1) }}} 

We know  {{{ W = 2^(k) + (-1)^(k+1) }}} is divisible by 3 (by hypothesis)
<br>

Look at V-W:   
{{{ V-W = 2^(k+1)+(-1)^((k+1)+1) - 2^k - (-1)^(k+1) }}}<br>
 = {{{ 2^(k+1)-2^k + ((-1)^((k+1)+1) - (-1)^(k+1)) }}} <br>
The last two terms combine to -2 when k is odd, +2 when k is even, factor
out 2:
 = {{{ 2(2^(k)-2^(k-1) + (-1)^(k)) }}}<br>
Noting that {{{2^k-2^(k-1) = 2^(k-1) }}}:
 = {{{ 2 * green( ( 2^(k-1) + (-1)^k) ) }}} <br>

By the hypothesis, the {{{ green(green) }}} factor is divisible by 3 

We know W is divisible by 3 and V-W is also divisible by 3, so V = (W)+(V-W) is also divisible by 3.<br>

   DONE