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
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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
(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](/uploads/image/z/6543585-9-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3Da%2CS%28n%2B1%29%3D2Sn%2Bn%2B1+%281%29%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9+%282%29%E5%BD%93a%3D1%E6%97%B6%2C%E8%8B%A5bn%3Dn)
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!
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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