Math
posted by K .
Use mathematical induction to prove the property for all positive integers n.
A factor of (n^3+3n^2+2n) is 3.

Math 
Steve
1^3+3*1^2+2*1 = 1+3+2 = 6 = 3*2
so, true for n=1
so, assuming it's true for n=k, work with n=k+1:
(k+1)^3 + 3(k+1)^2 + 2(k+1)
= k^3+3k^2+3k+1 + 3k^2+6k+3 + 2k+2
= k^3+3k^2+2k + 3k+1 + 3k^2+6k+3 + 2
we know that k^3+3k^2+2k is a multiple of 3, so check out the rest. Rearranging things, we have
3k^2+9k+3
This is clearly also a multiple of 3.
So,
P(1)
P(k) => P(k+1)
so P is true for all k.
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