n=2，3^2=9>； 8=2^3

n=3,3^3=27=3^3

2. if n=k, 3k >；; k^3(k & gt； 3)

Then when n=k+ 1

x^(k+ 1)=3*3^k>； =3k^3

Suppose 3k 3 > = (k+1) 3, then x (k+1) > = (k+1) 3.

According to mathematical induction, 3n >；; =n^3

And 3k 3 > = (k+ 1) 3, that is, 3k3 > = k 3+3k 2+3k+1,2k 3-3k 2-3k-1> =0 ....( 1)

Let f (x) = 2x3-3k2-3x-1(x >; 1, x is an integer)

f'(x)=6x^2-6x-3=6(x^2-x- 1/2)=6[(x- 1/2)^2-3/4]

Obviously f'(x) is increasing function, when x >; At 1, f' (x) > = f (2) = 6 [(2-1/2) 2-3/4] = 6 [9/4-3/4] = 9 > 0

F'(x)>0, and f(x) is also increasing function.

f( 1)= 2-3-3- 1 =-6 & lt； 0

f(2)=2*2^3-3*2^2-3*2- 1= 16- 12-6- 1=-3<； 0

f(3)=2*3^3-3*3^2-3*3- 1=54-27-9- 1= 17>； 0

At X>2 o'clock, f(x)>0 holds.

Therefore, point k>2, (1) holds.

K>3 is set according to the topic, so (1) holds.

So the hypothesis is true.

Therefore, 3 n > = n 3, this equation only holds when n=3.