Darrell K. answered • 06/14/18

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Experienced High School and College Tutor

**Step 1:**Test n=1(base case), we get (1)(2)(3)=6=3x2; true, then Distribute terms

**Step 2:**Assume true for n=k, k

^{3}+3k

^{2}+2k=3*P, where k and P are positive integers if you

**Step 3:**Check the successor to k, n=k+1, (k+1)((k+1+1)((k+1)+2)=3Q, (k+1)(k+2)(k+3)=3Q.

Distribute and we get k

^{3}+3k^{2}+3k+1+3k^{2}+6k+3+2k+2=3QRegroup terms (k

^{3}+3k^{2}+2k)+3k+1+3k^{2}+6k+3+2=3. Remember (k^{3}+3k^{2}+2k)=3P so, substitute3P+3k

^{2}+9k+6=3Q. I am going to group terms, (3P+6)+(3k^{2}+9k)=3Q, factor out a 3 on both expressions**3(P+2)+3k(k+3)=3Q**from there it is obvious to see that this is also divisible by 3.Proved by induction