Question 1102028
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Show by	the Principle of Mathematical Induction:

	
8 + 16 + 24 + . . . + 8n = 4n(n+1)     for  n >= 1.          (1)



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1.  In accordance with the Principle of Mathematical Induction, we should first check validity of the statement at n= 1.

    At n= 1 the left side is equal to 8;  the right side is 4*1*(1+1) = 4*2 = 8, too,

    and the statement is valid.



2.  Now assume that the statement is valid at some integer n= k:

    8 + 16 + 24 + . . . + 8k = 4k*(k+1).     (2)


    Then 

    8 + 16 + 24 + . . . + 8k + 8*(k+1) = [8 + 16 + 24 + . . . + 8k] + 8*(k+1) =  (replacing the brackets [ ] by  (1) )  = 


                                       = 4k*(k+1) + 8*(k+1) = (k+1)*(4k+8) = 4*((k+1)*(k+2).    (3)


    Thus you see (and I proved it) that if  the formula (1) is valid at n= k:

        8 + 16 + 24 + . . . + 8k = 4k*(k+1),

    then  the same formula (1) is valid for the next value of n= k+1:

       8 + 16 + 24 + . . . + 8k + 8*(k+1)  =  4*(k+1)*(k+2) = 4*(k+1)*((k+1)+1).



3.  According to the Principle of Mathematical Induction, it means that the formula (1) is valid for all integer n >= 1.
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QED.