设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1
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设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1
=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;
=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;
![设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1](/uploads/image/z/17468002-10-2.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B3Sn-tSn-1%3Dn%28n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%2Cn%E5%B1%9E%E4%BA%8EN%29%2Ct%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E4%B8%94a1)
1.当t=2时,取n=2有S2-2S1=2即a1+a2-2a1=2,又a1=1得a2=3;取n=3时有a1+a2+a3-2a1-2a2=3带入得a3=7
2.由Sn-tSn-1=n,a1=1得a2=1+t,a3=1+2t+t^2,因为an +1是等比,带入(a2 +1)^2=(a1+1)*(a3+1)得t^2=0,即t=0,an恒为1
2.由Sn-tSn-1=n,a1=1得a2=1+t,a3=1+2t+t^2,因为an +1是等比,带入(a2 +1)^2=(a1+1)*(a3+1)得t^2=0,即t=0,an恒为1
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