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已知数列{an}的各项均为正数,它的前n项和Sn满足Sn
1
6
(an+1)(an+2)
,并且a2,a4,a9成等比数列.
(1)求数列{an}的通项公式;
(2)设bn=(-1)n+1anan+1,Tn为数列{bn}的前n项和,求T2n
人气:428 ℃ 时间:2020-03-17 20:30:47
解答
(1)∵对任意n∈N*,有Sn=16(an+1)(an+2)①当n≥2时,有Sn−1=16(an−1+1)(an−1+2)②当①-②并整理得(an+an-1)(an-an-1-3)=0,而{an}的各项均为正数,所以an-an-1=3.∴当n=1时,有S1=a1=16(a1+1)(a1+2),...
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