已知數(shù)列{an}的前n項(xiàng)和Sn=12n2+12n(n∈N*),{bn}是公比大于0的等比數(shù)列,且滿足b1=a3,b2+b3=36.
(Ⅰ)求{an}和{bn}的通項(xiàng)公式;
(Ⅱ)若數(shù)列{1a2n-1a2n+1}的前n項(xiàng)和為Tn(n∈N*),求證:13≤Tn<12;
(Ⅲ)對(duì)任意的正整數(shù)n,設(shè)數(shù)列cn=an,n為奇數(shù), bn,n為偶數(shù).
求n∑k=1c2k+1c2k.
1
2
n
2
+
1
2
n
(
n
∈
N
*
)
{
1
a
2
n
-
1
a
2
n
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1
}
1
3
≤
T
n
<
1
2
a n , n 為奇數(shù) , |
b n , n 為偶數(shù) . |
n
∑
k
=
1
c
2
k
+
1
c
2
k
【考點(diǎn)】錯(cuò)位相減法.
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【解答】
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發(fā)布:2024/6/27 10:35:59組卷:265引用:1難度:0.6
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