已知数列{an},{bn}分别是等差、等比数列,且a1=b1=1,a2=b2,a4=b3≠b4.①求数列{an},{bn}的通项公式;②设Sn为数列{an}的前n项和,求{1Sn}的前n项和Tn;③设

题目简介

已知数列{an},{bn}分别是等差、等比数列,且a1=b1=1,a2=b2,a4=b3≠b4.①求数列{an},{bn}的通项公式;②设Sn为数列{an}的前n项和,求{1Sn}的前n项和Tn;③设

题目详情

已知数列{an},{bn}分别是等差、等比数列,且a1=b1=1,a2=b2,a4=b3≠b4
①求数列{an},{bn}的通项公式;
②设Sn为数列{an}的前n项和,求{
1
Sn
}的前n项和Tn
③设Cn=
anbn
Sn+1
(n∈N),Rn=C1+C2+…+Cn,求Rn
题型:解答题难度:中档来源:不详

答案

①设{an}的公差为d,{bn}的公比为q,则依题意
1+d=q
1+3d=q2
q≠1
q=2
d=1

∴an=1+(n-1)×1=n;
bn=1×2n-1=2n-1.(4分)
②∵sn=
n(n+1)
2
class="stub"1
sn
=class="stub"2
n(n+1)
=2(class="stub"1
n
-class="stub"1
n+1
).
∴Tn=class="stub"1
s1
+class="stub"1
s2
+…+class="stub"1
sn

=2[(class="stub"1
1
-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)]
=2(1-class="stub"1
n+1

=class="stub"2n
n+1
.(8分)
③∵Cn=
n•2n-1
(n+1)(n+2)
2
=
n•2n
(n+1)(n+2)
=
2n+1
n+2
-
2n
n+1

∴Rn=C1+C2+…+Cn
=(
22
3
-
21
2
)+(
23
4
-
22
3
)+…+(
2n+1
n+2
-
2n
n+1

=
2n+1
n+2
-1.

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