(20 14? Wuhu simulation) It is known that the side length of positive △ABC is 3, P 1 is a point on the side of AB, and BP 1= 1. From P 1 to BC, the vertiCAl foot is Q 1, and from Q 1 to ca.

Then bqn = bpncos 60 = 12an,

QnC=3？ 12an，

CRn=QnCcos60 = 12(3？ 12an)，

ARn=3？ CRn=32+ 14an，

APn+ 1 = arn cos 60 = 12 arn = 34+ 18an，

∴BPn+ 1=3？ APn+ 1=94？ 18an，

That is an+ 1 =? 18an+94？ (n∈N*)。

(2) From an+ 1 =? 18an+94？ (n∈N*)，get: an+ 1？ 2=? 18 (Ann? 2),

∴{an-2} Based on a 1-2=- 1, what is the common ratio? Geometric series of 18.

∴an？ 2=? (? 18)n？ 1, which means an = 2? (? 18)n？ 1? (n∈N*)。

∴nan=2n？ n(？ 18)n？ Then 1

Sn=2( 1+2+3+…+n)？ [ 1? (? 18)0+2? (? 18) 1+…+n(？ 18)n？ 1],

Make TN = 1? (? 18)0+2? (? 18) 1+…+n(？ 18)n？ 1,

18Tn= 1？ (? 18) 1+2? (? 18)2+…+n？ (? 18)n，

Subtract the two expressions to get:

98Tn= 1+(？ 18)+(? 18)2+…+(? 18)n？ 1? n(？ 18)n= 1？ (? 18)n 1？ (? 18)? n(？ 18)n，

∴Tn=64+(9n+8)(？ 18)n？ 18 1.

∴Sn=n(n+ 1)？ (9n+8)(？ 18)n？ 18 1.