數(shù)列{an}的前項(xiàng)和為Sn.已知a1=1,an+1=Sn.求證數(shù)列{}是等比數(shù)列.且Sn+1=4an 查看更多

 

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設(shè)數(shù)列{an}的前項(xiàng)和為Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*).

(1)求a1,a2的值;

(2)求證:數(shù)列{Sn+2}是等比數(shù)列;

(3)抽去數(shù)列{an}中的第1項(xiàng),第4項(xiàng),第7項(xiàng),……,第3n-2項(xiàng),……,余下的項(xiàng)順序不變,組成一個(gè)新數(shù)列{bn},若{bn}的前n項(xiàng)的和為T(mén)n,求證:

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設(shè)數(shù)列{an}前n項(xiàng)和為Sn,已知a1=a(a≠4),an+1=2Sn+4n(n∈N*
(Ⅰ)設(shè)b n=Sn-4n,求證:數(shù)列{bn}是等比數(shù)列;
(Ⅱ)求數(shù)列{an}的通項(xiàng)公式;
(Ⅲ)若an+1≥an(n∈N*),求實(shí)數(shù)a取值范圍.

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設(shè)數(shù)列{an}前n項(xiàng)和為Sn,已知a1=a(a≠4),an+1=2Sn+4n(n∈N*
(Ⅰ)設(shè)b n=Sn-4n,求證:數(shù)列{bn}是等比數(shù)列;
(Ⅱ)求數(shù)列{an}的通項(xiàng)公式;
(Ⅲ)若an+1≥an(n∈N*),求實(shí)數(shù)a取值范圍.

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設(shè)數(shù)列{an}前n項(xiàng)和為Sn,已知a1=a(a≠4),an+1=2Sn+4n(n∈N*
(Ⅰ)設(shè)b,求證:數(shù)列{bn}是等比數(shù)列;
(Ⅱ)求數(shù)列{an}的通項(xiàng)公式;
(Ⅲ)若an+1≥an(n∈N*),求實(shí)數(shù)a取值范圍.

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數(shù)列{an}前n項(xiàng)和為Sn,且Sn=an2+bn+c(a,b,c∈R),已知a1=-28,S2=-52,S5=-100.
(1)求數(shù)列{an}的通項(xiàng)公式.
(2)求使得Sn最小的序號(hào)n的值.

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