数列{an}满足a1=1/6,前n项和Sn=n(1/2)(n+1)an(1)写出S1,S2,S3,S4,S5,并由此猜出Sn的表达式(2)并用数列归纳法证明你的结论
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![数列{an}满足a1=1/6,前n项和Sn=n(1/2)(n+1)an(1)写出S1,S2,S3,S4,S5,并由此猜出Sn的表达式(2)并用数列归纳法证明你的结论](/uploads/image/z/1159362-18-2.jpg?t=%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3D1%2F6%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3Dn%281%2F2%29%28n%2B1%29an%281%29%E5%86%99%E5%87%BAS1%2CS2%2CS3%2CS4%2CS5%2C%E5%B9%B6%E7%94%B1%E6%AD%A4%E7%8C%9C%E5%87%BASn%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%282%29%E5%B9%B6%E7%94%A8%E6%95%B0%E5%88%97%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%E4%BD%A0%E7%9A%84%E7%BB%93%E8%AE%BA)
数列{an}满足a1=1/6,前n项和Sn=n(1/2)(n+1)an(1)写出S1,S2,S3,S4,S5,并由此猜出Sn的表达式(2)并用数列归纳法证明你的结论
数列{an}满足a1=1/6,前n项和Sn=n(1/2)(n+1)an
(1)写出S1,S2,S3,S4,S5,并由此猜出Sn的表达式
(2)并用数列归纳法证明你的结论
数列{an}满足a1=1/6,前n项和Sn=n(1/2)(n+1)an(1)写出S1,S2,S3,S4,S5,并由此猜出Sn的表达式(2)并用数列归纳法证明你的结论
1) Sn=n(n+1)/2*an
S(n-1)=(n-1)n/2*a(n-1)
两式相减得∶Sn-S(n-1)=an=n(n+1)/2*an-(n-1)n/2*a(n-1)
整理得an/a(n-1)=n/(n+2)
用累积法求出an=1/(n+1)(n+2)
Sn=n/(2n+4)
2)(A)当n=1时S1=1/(2+4)=1/6=a1满足条件
(B)假设当n=k时,Sn=n/(2n+4)成立
则Sk=k/(2k+4)则当n=k+1时,
S(k+1)=Sk+a(k+1)= k/(2k+4)+2S(n+1)/(n+1)(n+2)
整理得S(k+1)=(k+1)/(2(k+1)+4)
由A、B可证得Sn=n/(2n+4)
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