摘要：解:(1)∵a10=5,d=2,∴an=2n-15 又∵b3=4,q=2,∴bn=2n-1∴cn=?2n-1(2)Sn=c1+c2+c3+-+cn,2Sn=2c1+2c2+2c3+-+2cn错位相减.得-Sn=c1+(c2-2c1)+(c3-2c2)+-+(cn-2cn-1)-2cn∵c1=-13,cn-2cn-1=2n∴-Sn=-13+22+23+-+2n-?2n=-13+4(2n-1-1)-?2n=-17+2n+1-?2n ∴Sn=17+?2n∴==.

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