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

Asked by Theodosia Lourdes | 8th May, 2011, 03:53: AM

Expert Answer:

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.

Answered by  | 8th May, 2011, 10:27: AM

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