Try this for n = 1:

1*3 =? 1(1 + 1)(2*1 +7)/6

3 =? 1*2*9/6

3 =? 3 (Check!)

Now assume the statement is true for some integer k. That is

1*3 + 2*4 + ... + k(k + 2) = k(k + 1)(2k + 7)/6

We need to show that it also holds for k + 1:

1*3 + 2*4 + ... + k(k + 2) + (k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)/6

We know that 1*3 + ... + k(k + 2) = k(k + 1)(2k + 7)/6 so use that in the above equation. Thus

k(k + 1)(2k + 7)/6 + (k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)/6

should be an identity.

Multiply both sides by 6, then simplify:

k(k + 1)(2k + 7) + 6(k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)

(k^2 + k)(2k + 7) + 6(k^2 + 4k + 3) = (k^2 + 3k + 2)(2k + 9)

2k^3 + 9k^2 + 7k + 6k^2 + 24k + 18 = 2k^3 + 15k^2 + 31k + 18

2k^2 + 15k^2 + 31k + 18 = 2k^3 + 15k^2 + 31k + 18

0 = 0 (Check!)

Thus the proposition is true for k = 1, so it is true for all integer k >= 1.

-Dan