(1)由条件求得an=2n-15,bn=2n-1,可得cn =anbn=(2n-15)•2n-1.
(2)根据 Sn=c1+c2+c3+…+cn,用错位相减法求得错位相减求得Sn的解析式,再利用数列极限的运算法则求得= 的值.
【解析】
(1)∵a10=5,d=2,∴an=2n-15.
又∵b3=4,q=2,∴bn=2n-1,∴cn=(2n-15)•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-15)•2n=-13+4(2n-1-1)-(2n-15)•2n
=-17+2n+1-(2n-15)•2n∴Sn=17+(2n-17)•2n.
∴=
= .