已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),2a5,a4,4a6成等差數(shù)列,且滿足a4=4a23,等差數(shù)列數(shù)列{bn}的前n項(xiàng)和Sn,b2+b4=6,S4=10
(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;
(2)設(shè)cn=bn(n為奇數(shù)) an?bn(n為偶數(shù))
,求數(shù)列{cn}的前2n項(xiàng)和.
(3)設(shè)dn=b2n+5b2n+1b2n+3an,n∈N*,{dn}的前n項(xiàng)和Tn,求證:Tn<13.
a
4
=
4
a
2
3
c
n
=
b n ( n 為奇數(shù) ) |
a n ? b n ( n 為偶數(shù) ) |
d
n
=
b
2
n
+
5
b
2
n
+
1
b
2
n
+
3
a
n
,
n
∈
N
*
,
{
d
n
}
T
n
<
1
3
【考點(diǎn)】錯(cuò)位相減法.
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發(fā)布:2024/6/27 10:35:59組卷:343引用:1難度:0.5
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