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(2011•天津模擬)設(shè)數(shù)列{an}滿足a1=a,an+1=can+1-c(n∈N*),其中a,c為實(shí)數(shù),且c≠0.
(1)求證:a≠1時(shí)數(shù)列{an-1}是等比數(shù)列,并求an
(2)設(shè)a=
1
2
c=
1
2
,bn=n(1-an)(n∈N*)
,求數(shù)列{bn}的前n項(xiàng)和Sn;
(3)設(shè)a=
3
4
,c=-
1
4
cn=
3+an
2-an
(n∈N*),記dn=c2n-c2n-1(n∈N*)
,設(shè)數(shù)列{dn}的前n項(xiàng)和為T(mén)n,求證:對(duì)任意正整數(shù)n都有Tn
5
3

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在數(shù)列{an}中,an+1=can(c為非零常數(shù))且前n項(xiàng)和Sn=3n+k,則k等于( 。

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已知數(shù)列{an}中,a1=1,an=can-1+2n(n≥2,c是大于0的常數(shù)),且a3=13.
(1)求c的值;   
(2)求{an}的通項(xiàng)公式.

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在數(shù)列{an}中,an+1=can(c為非零常數(shù)),且其前n項(xiàng)和為Sn=3n+k,則實(shí)數(shù)k+c的值為
2
2

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等比數(shù)列ca1,ca2,ca3,…,can(c為非零常數(shù))公比為-2,數(shù)列a1,a2,a3,…,an是(  )

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