Required to prove n^3 + 3n^2 + 2n is divisible by 6.
Using induction:
Let S(n) = n^3 + 3n^2 + 2n where n is an element of N
Now, S(1) = 1+3+2 = 6
Thus S(1) is divisible by 6 (i.e. statement true for n = 1)
Assume that S(k) = k^3 + 3k^2 + 2k = 6M where M is an element of Z
Now, S(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^2 + 9k + 6
= 6M + 3k^2 + 9k + 6 (by the induction hypothesis)
= 6M + 3(k+1)(k+2)
Now if k+1 is odd, then k+2 is even and thus k+2 = 2L where L is an element of Z.
If k+1 is even, then k+1 = 2N where N is an element of Z.
Thus, S(k+1) = 6M + 3(2N)(k+2) = 6{M+N(k+2)}
or S(k+1) = 6M + 3(2L)(k+1) = 6{M+L(k+1)}
In either case, S(k+1) is divisible by 6.
Thus, the statement is true for n=1 and if it is true for n=k then it is true for n=k+1. Thus by the principle of mathematical induction, the statement is true for n=1,2,3,.... for n an element of N.
Without induction (by factorising)
n^3 + 3n^2 + 2n = n(n^2+3n+2) = n(n+1)(n+2)
If n is even, then the statement is divisible by 2.
If n is odd, then n+1 is even and still the statement is divisible by 2.
Thus the statement is divisible by 2 for all n an element of N.
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Multiples of 3 include 3,6,9,.... forming an arithmetic progression with common difference 3. The k-th term is thus given by T_n = 3k.
Case 1: n=3k+2
Then, n+1=3k+3=3(k+1).
Thus the statement becomes n*3(k+1)(n+2) and because n+1 is odd, both n and n+2 are even. Thus since the statement is divisible by both 2 and 3, it is divisible by 6.
Case 2: n=3k+1
Then, n+2=3k+3=3(k+1)
Thus the statement becomes n(n+1)*3(k+1) and because n+2 is odd, n+1 is even. Thus since the statement is divisible by both 2 and 3, it is divisible by 6.
Case 3: n=3k
Thus the statement becomes 3k(n+1)(n+2) and because n is odd, n+1 is even. Thus since the statement is divisible by both 2 and 3, it is divisible by 6.
In all three cases n=3k,3k+1,3k+2, the statement is divisible by 6.
Thus the statement is divisible by 6 for all n an element of N.
