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设正数数列{an}为一等比数列,且a2=4,a4=16.求:
lim
n→∞
lgan+1+lgan+2+…+lga2n
n
人气:279 ℃ 时间:2020-05-03 08:27:05
解答
设数列{an}的公比为q,显然q≠1,
a4
a2
q2=4
,由于an>0,n∈N,
∴q=2,a1
a2
q
=2
,∴an=a1qn-1=2n
因此
lgan+1+lgan+2+…+lga2n
n
=
lg2n+1+lg2n+2+…+lg22n
n2

=
[(n+1)+(n+2)+…+2n]
n2
lg2

=
3n2+n
2n2
•lg2

原式=
lim
n→∞
(
3n2+n
2n2
•lg2) =lg2•
lim
n→∞
3n2+n
2n2
=
3
2
lg2
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