已知:等比数列{an}中,a1=3,a4=81,(n∈N*).(1)若{bn}为等差数列,且满足b2=a1,b5=a2,求数列{bn}的通项公式;(2)若数列{bn}满足bn=log3an,求数列{1

题目简介

已知:等比数列{an}中,a1=3,a4=81,(n∈N*).(1)若{bn}为等差数列,且满足b2=a1,b5=a2,求数列{bn}的通项公式;(2)若数列{bn}满足bn=log3an,求数列{1

题目详情

已知:等比数列{an}中,a1=3,a4=81,(n∈N*).
(1)若{bn}为等差数列,且满足b2=a1,b5=a2,求数列{bn}的通项公式;
(2)若数列{bn}满足bn=log3an,求数列{
1
bnbn+1
}
的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)在等比数列{an}中,a1=3,a4=81.
所以,由a4=a1q3得3q3=81,
解得q=3.
因此,an=3×3n-1=3n.在等差数列{bn}中,
根据题意,b2=a1=3,b5=a2=9,,可得,
d=
b5-b2
5-2
=2
所以,bn=b2+(n-2)d=2n-1
(Ⅱ)若数列{bn}满足bn=log3an,
则bn=log33n=n,
因此有class="stub"1
b1b2
+class="stub"1
b3b2
+…+class="stub"1
bnbn+1
=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)=class="stub"n
n+1

更多内容推荐