n(n+1)(n+2) is divisible by 3

Question

Mathematical induction. Any solution as a worked example would be greatly appreciated. Thanks

Answer 1

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, k³+3k²+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 k³+3k²+3k+1+3k²+6k+3+2k+2=3Q

Regroup terms (k³+3k²+2k)+3k+1+3k²+6k+3+2=3. Remember (k³+3k²+2k)=3P so, substitute

 

3P+3k²+9k+6=3Q. I am going to group terms, (3P+6)+(3k²+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

Answer 2

Empirically, you can see that whenever you list three consecutive positive integer, exactly one of them must be divisible by 3.

 

Assume n is not divisible by 3, so let's call it 3m+1. That is, assume it's one more than a multiple of three.

Then we have

 

n     =     3m+1

n+1 =    3m +2

n+2 =    3m+3

Thus it is evident that n+2 is divisible by 3 if n = 3m+1.

 

Now do a similar study assuming that n = 3m+2; in that case n+1 is divisible by 3.

 

The only other case is that n itself is a multiple of 3.

 

Thus one can see that exactly one of the numbers n, n+1 & n+2 has to be a multiple of 3.