题目
已知公差为d(d>1)的等差数列{an}和公比为q(q>1)的等比数列{bn},满足集合{a3,a4,a5}∪{b3,b4,b5}={1,2,3,4,5}, (1)求通项an,bn; (2)求数列{an·bn}的前n项和Sn.
答案:(1)∵1,2,3,4,5这5个数中成公差大于1的等差数列的三个数只能是1,3,5;成公比大于1的等比数列的三个数只能是1,2,4. 而{a3,a4,a5}∪{b3,b4,b5}={1,2,3,4,5}, ∴a3=1,a4=3,a5=5,b3=1,b4=2,b5=4, ∴a1=-3,d=2,b1=,q=2, ∴an=a1+(n-1)d=2n-5,bn=b1×qn-1=2n-3. (2)∵anbn=(2n-5)×2n-3, ∴Sn=(-3)×2-2+(-1)×2-1+1×20+…+(2n-5)×2n-3, 2Sn=-3×2-1+(-1)×20+…+(2n-7)×2n-3+(2n-5)×2n-2, 两式相减得-Sn=(-3)×2-2+2×2-1+2×20+…+2×2n-3-(2n-5)×2n-2=--1+2n-1-(2n-5)×2n-2 ∴Sn=+(2n-7)×2n-2.