Using the principle of mathematical induction prove that for all n belongs to N: 6 power"n+2" + 7 power"2n+1" is divisible by 43

Let P(n) be the statement ( 6 ⁽ⁿ⁺²⁾+ 7⁽²ⁿ⁺¹⁾) = 43x.

1. Basis of Induction

            Since ( 6(1+2)+ 7(2+1)) = 63  + 73  = 559 = 43 X 13, 
   the formula is true for n = 1.
   		

2. Inductive Hypothesis

                                                
   Assume that P(n) is true for n = k, that is 6(k+1)+ 7(2k+1)) = 43x 
   for some integer x.
   		

3. Inductive Step

   Now show that the formula is true for n = k + 1. 
   
   Observe that 
             
   P(k+1) = 6(k+1+2)+ 7(2(k+1)+1)
   
             = 6(k+3)+ 7(2k+3)
   
                 
          = 6 X 6(k+2)+ 72  X 7(2k+1)
   
                 
          = 6 X 6(k+2)+(6+43) X 7(2k+1)
   
                 
          = 6 X 6(k+2)+ 6 X 7(2k+1)+ 43 X 7(2k+1)
   
                   
          = 6 X ( 6(k+2)+ 7(2k+1) )+ 43 X 7(2k+1)
   
                                 
          = 6 X ( P(k) ) + 43 X 7(2k+1)
   	

   Since each component of this sum is divisible by 43 so is the entire sum and the formula holds for k + 1.