Question 1135717

n=1:  (3*1)*(1+1) = 6
n=2:  (3*2)*(2+1) = 18 = 6 + 12
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Assume true for n=k:  n=k --> Sum = 3k(k+1)
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Now let n=k+1:

  Sum = (6+12+18+...+6k) + 6(k+1)
      = 3k(k+1) + 6(k+1)
      = 3(k+1)(k+2)    <<<< factored out 3(k+1) from previous line

This concludes the proof.