Proof of 6|(-2/3)*{1+(-2)^n-2^(2*n+1)} by method of induction
                                    ----Indranil Ghosh, 17th January, 2017.
This problem can be written as:
             Proof of 9|-{1+(-2)^n-2^(2*n+1)} by induction
           =>Proof of 9|{2^(2*n+1)-(-2)^n-1}........................................(1)
Step 1: For n=1, (1) is true, since
                              2^(2*1+1)-(-2)^1-1
                             =2^3-(-2)-1
                             =9.
                             And, 9|9.
Step 2: Suppose (1) is true for some n = k >= 1, i.e.,
                             9|{2^(2*k+1)-(-2)^k-1}.
Step 3: Prove that (1) is true for n = k+1, i.e.,
                             9|[2^{2*(k+1)+1}-(-2)^(k+1)-1]
            We have,   2^{2*(k+1)+1}-(-2)^(k+1)-1
                     = 2^(2*k+3)-(-2)^(k+1)-1
                     = 4*{2^(2*k+1)}-(-2)*(-2)^k-1                             
                     = {2^(2*k+1)-(-2)^k-1}*4 + {6*(-2)^k+3}
                     = a1 + a2  [a1 = {2^(2*k+1)-(-2)^k-1}*4 and a2 = {6*(-2)^k+3}]
            Now, 9|a1  [As evident from step 2]
            For a2, we again apply the method of induction to proof that 9|a2.
             Proof of 9|6*(-2)^k+3..................................................(2)
Step 4: For k = 1  (2) is true since,
                     6*(-2)^1+3
                   = -12+3
                   = -9
                   And 9|-9
Step 5: Suppose (2) is true for k = m>=1 i.e.,
                   9|6*(-2)^m+3
 Step 6: Prove that (2) is true for k=m+1 i.e.,
                   9|6*(-2)^(m+1)+3
            We have  6*(-2)^(m+1)+3
                   = 6*{(-2)^m}*(-2)+3
                   = -12*(-2)^m+3
                   = {6*(-2)^m+3]*(-2)+9
                   And 9|{6*(-2)^m+3]*(-2)+9 [As evident from step 5]
            So,     9|a1 and 9|a2
                 => 9|(a1+a2)
	             => 9|[2^{2*(k+1)+1}-(-2)^(k+1)-1] (Proved)
So, 9|{2^(2*n+1)-(-2)^n-1}
=>9|-{1+(-2)^n-2^(2*n+1)}
=>6|(-2/3)*{1+(-2)^n-2^(2*n+1)}  (Proved)