已知{an}為等差數(shù)列,{bn}為公比大于0的等比數(shù)列,且b1=2,b2+b3=12,a3=3,a4+2a6=b4.
(1)求{an}和{bn}的通項(xiàng)公式;
(2)dn=(3an+1+5)bn+1(anbn+1)(an+2bn+2+1),n=2k-1 anbn,n=2k
,(k∈N*),求數(shù)列{dn}的前2n項(xiàng)和S2n;
(3)記Cm為{bn}在區(qū)間(0,m](m∈N*)中項(xiàng)的個(gè)數(shù),求數(shù)列{Cm}的前200項(xiàng)和T200.
d
n
=
( 3 a n + 1 + 5 ) b n + 1 ( a n b n + 1 ) ( a n + 2 b n + 2 + 1 ) , n = 2 k - 1 |
a n b n , n = 2 k |
【考點(diǎn)】錯(cuò)位相減法.
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發(fā)布:2024/6/27 10:35:59組卷:164引用:1難度:0.5
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