设数列的前n项和为Sn,已知a1+2a2+3a3+……+nan=(n-1)Sn+2n(n属于N+)(1)求a2,a3的值(2)求证{Sn+2}是等比数列
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![设数列的前n项和为Sn,已知a1+2a2+3a3+……+nan=(n-1)Sn+2n(n属于N+)(1)求a2,a3的值(2)求证{Sn+2}是等比数列](/uploads/image/z/1071848-56-8.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%2B2a2%2B3a3%2B%E2%80%A6%E2%80%A6%2Bnan%3D%28n-1%29Sn%2B2n%EF%BC%88n%E5%B1%9E%E4%BA%8EN%2B%EF%BC%89%281%29%E6%B1%82a2%2Ca3%E7%9A%84%E5%80%BC%282%29%E6%B1%82%E8%AF%81%7BSn%2B2%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97)
设数列的前n项和为Sn,已知a1+2a2+3a3+……+nan=(n-1)Sn+2n(n属于N+)(1)求a2,a3的值(2)求证{Sn+2}是等比数列
设数列的前n项和为Sn,已知a1+2a2+3a3+……+nan=(n-1)Sn+2n(n属于N+)
(1)求a2,a3的值
(2)求证{Sn+2}是等比数列
设数列的前n项和为Sn,已知a1+2a2+3a3+……+nan=(n-1)Sn+2n(n属于N+)(1)求a2,a3的值(2)求证{Sn+2}是等比数列
(1).令n=1,得:a1=2
令n=2,也易得a2=4
令n=3,可得a3=8
(2).
已知a1+2a2+3a3+……+nan=(n-1)Sn+2n.①
则 a1+2a2+3a3+……+nan+(n+1)a(n+1)=nS(n+1)+2(n+1).②
②-①得:
(n+1)a(n+1)=nS(n+1)-nSn+Sn+2.③
又 a(n+1)=S(n+1)-Sn
∴③可变型为:
(n+1)(S(n+1)-Sn)=nS(n+1)-nSn+Sn+2
化简得:
S(n+1)=2Sn+2
等号两边同时加2,得:
S(n+1)+2=2(Sn+2)
( S(n+1)+2)/(Sn+2)=2
故:{Sn+2}是以S1+2=2+2=4为首项,2为公比的等比数列
过程中有不懂得可以在线问我
a1=(1-1)Sn+2,a1=2
a1+2a2=S2+4=a1+a2+4,a2=4
同理将a3带入得,a3=8
a1+2a2+3a3+……+nan=(n-1)Sn+2n
a1+2a2+3a3+……+(n+1)a(n+1)=nS(n+1)+2(n+1)
两式相减得(n+1)a(n+1)=nS(n+1)-(n-1)Sn+2
(n+1)a(n+1)=n{...
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a1=(1-1)Sn+2,a1=2
a1+2a2=S2+4=a1+a2+4,a2=4
同理将a3带入得,a3=8
a1+2a2+3a3+……+nan=(n-1)Sn+2n
a1+2a2+3a3+……+(n+1)a(n+1)=nS(n+1)+2(n+1)
两式相减得(n+1)a(n+1)=nS(n+1)-(n-1)Sn+2
(n+1)a(n+1)=n{S(n+1)-Sn}+Sn+2
a(n+1)=Sn+2
S(n+1)-Sn=Sn+2
S(n+1)+2=2(Sn+2)
所以{Sn+2}是以2为公比,4为首的等比数列
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