设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ

题目简介

设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ

题目详情

设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=1,S5=45,T3=a3-b2
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)求
q
a1a2
+
q
a2a3
+…+
q
anan+1
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设{an}的公差为d,则S5=5+10d=45.             
解得d=4,所以an=4n-3.                 …(4分)
由T3=a3-b2,得1+q+q2=9-q,又q>0,从而解得q=2,所以bn=2n-1.       …(8分)
(Ⅱ)class="stub"q
anan+1
=class="stub"2
anan+1
=class="stub"d
2anan+1
=class="stub"1
2
(class="stub"1
an
-class="stub"1
an+1
)
.  …(10分)
所以M=class="stub"q
a1a2
+class="stub"q
a2a3
+…+class="stub"q
anan+1
=class="stub"1
2
(class="stub"1
a1
-class="stub"1
a2
+class="stub"1
a2
-class="stub"1
a3
+…+class="stub"1
an
-class="stub"1
an+1
)

=class="stub"1
2
(class="stub"1
a1
-class="stub"1
an+1
)
=class="stub"1
2
(1-class="stub"1
4n+1
)
=class="stub"2n
4n+1
.                           …(14分)

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