lishou8341
幼苗
共回答了22个问题采纳率:90.9% 举报
(Ⅰ)∵正项等比数列{an}满足:lna1+lna2=4,lna4+lna5=10,
∴a1a3=e4,a4a6=e10,
∴q6=e6,由q>0,解得q=e,a1=e,
∴an=en.
(Ⅱ)由(Ⅰ)知Sn=1+2+3+…+n=
n(n+1)
2,
bn=[1
n(n+1)=
1/n−
1
n+1],
∴b1+b2+…+bn=1-[1/2+
1
2−
1
3]+…+[1/n−
1
n+1]
=1-[1/n+1]
=[n/n+1],
设cn=(b1+b2+…+bn)([2/3])n,
∴cn=
n
n+1(
2
3)n,
cn+1-cn=[n+1/n+2]([2/3])n+1-[n/n+1]([2/3])n
=
−n2−2n+2
3(n+1)(n+2)•(
2
3)n<0,
∴cn>cn+1,
∴数列{cn}单调递减,
(cn)max=c2=[1/3],
∴k<[1/3].
1年前
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