(1) The chess piece is at point A on the upper and lower surfaces. If you move n times, the chess piece will fall on the vertex of the upper and lower surfaces. Starting from A, there are three paths, so P 1 = 23.

When the chess piece moves twice and is still up and down, there are two possibilities, p2=23×23+ 13( 1? 23)=59.

(2) Because after moving n times, the probability that the chess pieces fall on the upper and lower vertices is pn.

So the probability of falling on the vertex of the bottom surface is 1-pn.

Therefore, after moving n+ 1 times, the probability that the chess piece falls on the vertex of the upper bottom surface is pn+1= 23pn+13 (1? pn)= 13pn+ 13

, so that pn+1-12 =13 (pn? 12),

So the sequence {pn? 12} is a geometric series, the first term is 16, and the common ratio is 13, so pn? 12= 16×( 13)n？ 1,

Prove by mathematical induction: Ni = 1 14pi? 1>n2n+ 1。

(1) when n= 1, the left formula = 14×23? 1 = 35, and the formula on the right = 12, because 35 > 12, the inequality holds.

When n=2, the left formula = 14×23? 1+ 14×59? 1 = 7855, right formula =43, so the inequality holds;

② Suppose the inequality of n=k(k≥2) holds, that is, ki = 1 14pi? 1>k2k+ 1。

Then when n=k+ 1, the left formula = ki = 1 14pi? 1+ 14pk+ 1？ 1 > k2k+ 1+ 14( 12+ 12× 13k+ 1)？ 1 = k2k+ 1+3k+ 13k+ 1+2，

To prove that K2K+1+3k+13k+1+2 ≥ (k+1) 2k+2.

As long as the certificate is 3k+13k+1+2 ≥ (k+1) 2k+2? k2k+ 1，

It is proved that 3k+13k+1+2 ≥ k2+3k+1k2+3k+2,

As long as the certificate is 23k+1≤1K2+3k+1,

As long as the certificate is 3k+ 1≥2k2+6k+2,

Because k ≥ 2,3k+1= 3 (1+2) k ≥ 3 (1+2k+4c2k) = 6k2+3 = 2k2+6k+2+2k (2k-3)+1.

So k2k+1+3k+13k+1+2 ≥ (k+1) 2k+2,

That is, when n=k+ 1, the inequality also holds. From ① ②, we can know that Ni = 1 14pi? 1 > n2n+ 1 holds for any n∈N*.