设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数).求
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设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数).求证数列Sn+2是等比数列
![设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数).求](/uploads/image/z/7948174-22-4.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%A5a%281%29%2B2a%282%29%2B3a%283%29%2B%E2%80%A6%2Bna%28n%29%3D%28n-1%29Sn%2B2n%28n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89.%E6%B1%82)
由原式得a1+2a2+3a3```+nan=nSn-Sn+2n
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
∴nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2 (注意a1+2a2+3a3```+nan=nSn-Sn+2n
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2
就是通过2个式子的转化)
继续.
因为 S(n+1)-Sn=a(n+1)
-Sn+(n+1)a(n+1)=n(Sn+1-Sn)+2=na(n+1)+2
所以a(n+1)=Sn+2
∴an=Sn+2
∵ Sn-S(n-1)=an 上面2个式相减得到
a(n+1)=2an
∴an是公比为2的等比数列
由题可得a1=2 ∴Sn=2^(n+1)-2
即Sn+2=2^(n+1)
所以Sn+2 等比数列
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
∴nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2 (注意a1+2a2+3a3```+nan=nSn-Sn+2n
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2
就是通过2个式子的转化)
继续.
因为 S(n+1)-Sn=a(n+1)
-Sn+(n+1)a(n+1)=n(Sn+1-Sn)+2=na(n+1)+2
所以a(n+1)=Sn+2
∴an=Sn+2
∵ Sn-S(n-1)=an 上面2个式相减得到
a(n+1)=2an
∴an是公比为2的等比数列
由题可得a1=2 ∴Sn=2^(n+1)-2
即Sn+2=2^(n+1)
所以Sn+2 等比数列
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