1, when n= 1, 1×2×3=6, which is divisible by 6.

2. When n=2, 2×3×5=30, which is divisible by 6.

3. Suppose that when n=n- 1, (n- 1)n(2n- 1) is divisible by 6.

4, then when n = n, n(n+ 1)(2n+ 1).

= n[(n- 1)+2][(2n- 1)+2]

= n[(n- 1)(2n- 1)+2(n- 1)+2(2n- 1)+4]

=n(n- 1)(2n- 1)+n*6n

Because n(n- 1)(2n- 1) is divisible by 6, and 6n^2 is divisible by 6.

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2.

1. When n= 1, 1+8+27=36, which is divisible by 9.

2. When n=2, 8+27+64=99, which is divisible by 9.

3. Suppose that when n=n- 1, (n-1) 3+n 3+(n+1) 3 is divisible by 9.

4. So, when n=n,

n^3+(n+ 1)^3+(n+2)^3

= n^3+(n+ 1)^3+(n- 1+3)^3

=n^3+(n+ 1)^3+[(n- 1)^3+3*3(n- 1)^2+3*(n- 1)*3^2+3^3]

=n^3+(n+ 1)^3+(n- 1)^3+3*3(n- 1)^2+3*(n- 1)*3^2+3^3]

This equation is divisible by 9.

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