解（1）an＞0，a2=2，a4=8，∴q2＝ a4 a2 ＝ 8 2 ＝4，∴q=2则an＝a2?qn?2＝2n?1；（2）由（1）知b2=a3=4，b6=a5=16．∴d＝ b6?b2 6?2 ＝ 16?4 4 ＝3．d＝3，b1＝1，sn＝nb1+ n(n?1) 2 d＝ 3n2?n 2 ．
