Use mathematical induction to prove the statement is true for all positive integers n.6 + 12 + 18 + . . . + 6n = 3n(n + 1)

What will be an ideal response?


Step 1: Determine if S1, S2, and S3 are true. 
       
  
Step 2: Assume Sk to be true, then
   Sk: 6 + 12 + 18 + . . . + 6k = 3k(k + 1)

Step 3: Find Sk+1.  
   Sk+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 1 + 1)
   Sk+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 2)

Step 4: Truth of Sk implies that Sk+1 is true.  
   6 + 12 + 18 + . . . + 6k = 3k(k + 1)
 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k(k + 1) + 6(k + 1)
 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k2 + 3k + 6k + 6
 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k2 + 9k + 6
 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k2 + 3k + 2)
 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k + 1)(k + 2)

Since the last equation is Sk+1, it has been shown that the truth of Sk implies the truth of Sk+1 for every positive integer k. By the principle of mathematical induction, Sn is true for every positive integer n.

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