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bn>0(n∈N*),a1=b1=1,a2+b3=a3,S5=
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bn>0(n∈N*),a1=b1=1,a2+b3=a3,S5=5(T3+b2).(Ⅰ)求数列{a...
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bn>0(n∈N*),a1=b1=1,a2+b3=a3,S5=5(T3+b2).(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)求和:b1T1T2+b2T2T3+…+bnTnTn+1.
展开
1个回答
展开全部
(Ⅰ)设{an}的公差为d,数列{bn}的公比为q,则
∵a1=b1=1,a2+b3=a3,S5=5(T3+b2),
∴q2=d,1+2d=1+2q+q2,
∴q2-2q=0,
∵q≠0,∴q=2,∴d=4
∴an=4n-3,bn=2n-1;
(Ⅱ)∵
=
=
(
?
)
∴
+
+…+
=
(
?
+
?
+…+
?
)
=
(
?
)=
(1-
).
∵a1=b1=1,a2+b3=a3,S5=5(T3+b2),
∴q2=d,1+2d=1+2q+q2,
∴q2-2q=0,
∵q≠0,∴q=2,∴d=4
∴an=4n-3,bn=2n-1;
(Ⅱ)∵
bn |
TnTn+1 |
bn+1 |
qTnTn+1 |
1 |
q |
1 |
Tn |
1 |
Tn+1 |
∴
b1 |
T1T2 |
b2 |
T2T3 |
bn |
TnTn+1 |
1 |
q |
1 |
T1 |
1 |
T2 |
1 |
T2 |
1 |
T3 |
1 |
Tn |
1 |
Tn+1 |
=
1 |
q |
1 |
T1 |
1 |
Tn+1 |
1 |
q |
2 |
2n+1?1 |
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询