设数列an的前n项和为Sn,已知a1=a,S(n+1)=2Sn+n+1 (1)求an的通项 (2)当a=1时,若bn=n/[a(n+1)-an],数列b(2)当a=1时,若bn=n/[a(n+1)-an],数列bn的前n项和为Tn,证明Tn

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设数列an的前n项和为Sn,已知a1=a,S(n+1)=2Sn+n+1 (1)求an的通项 (2)当a=1时,若bn=n/[a(n+1)-an],数列b(2)当a=1时,若bn=n/[a(n+1)-an],数列bn的前n项和为Tn,证明Tn

设数列an的前n项和为Sn,已知a1=a,S(n+1)=2Sn+n+1 (1)求an的通项 (2)当a=1时,若bn=n/[a(n+1)-an],数列b(2)当a=1时,若bn=n/[a(n+1)-an],数列bn的前n项和为Tn,证明Tn
设数列an的前n项和为Sn,已知a1=a,S(n+1)=2Sn+n+1 (1)求an的通项 (2)当a=1时,若bn=n/[a(n+1)-an],数列b
(2)当a=1时,若bn=n/[a(n+1)-an],数列bn的前n项和为Tn,证明Tn

设数列an的前n项和为Sn,已知a1=a,S(n+1)=2Sn+n+1 (1)求an的通项 (2)当a=1时,若bn=n/[a(n+1)-an],数列b(2)当a=1时,若bn=n/[a(n+1)-an],数列bn的前n项和为Tn,证明Tn
S(n+1)= 2Sn + n +1 (1)
Sn = 2S(n-1) + (n -1)+1 (2)
(1)-(2)得:a(n+1)= 2an + 1 此等式左右两边同时加上1,
则可变成 a(n+1)+1= 2(an + 1 ) 则数列{a(n+1)+1}为等比数列
最后可求出an=2^(n-1)*(a+1)-1
第二问就迎刃而解了,第三问做不出来再追问吧!祝好!

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