Solution: (1) It is proved that when n= 1, S 1=b 1, b144d = b1(d+3d) 4d = b/kloc-0. ()

Assuming that when n=k, Sk=bkak+34d holds, (2 points)

Then sk+1= sk+bk+1= bkak+3+bk+14d4d (4 points).

= akak+ 1ak+2ak+3+bk+ 14d 4d = akbk+ 1+bk+ 14d 4d = bk+ 1(AK+4d)4d = bk+ 1ak。

So when n=k+ 1, the equation still holds, so for any n∈N*, there is Sn=b 1an+34d(8 points).

(2) because 3a5 = 8a 12 > 0, 3a5=8(a5+7d) and a5 =-56d5 > 0, d < 0.

And a16 = a5+1 1 d =-D5 > 0, a 17 = a5+ 12d = 4d5 < 0, (11.

So a1> A2 > A3 > … > A16 > 0 > A17 > A18, B1> B2 > B3 > … > B14 >.

Because b15 = a15a17a17 < 0, b16 = a16a17a18 > 0.

A 15 = a5+10d =-6d5 > 0, a 18=a5+ 13d=9d5 0, (15 points).

So S 16 > S 14, so S 16 is the largest in Sn. (16)