数列{an}的前n项和为Sn,已知a1+2,Sn+1=Sn-2nSn+1Sn,求an
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数列{an}的前n项和为Sn,已知a1+2,Sn+1=Sn-2nSn+1Sn,求an
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Sn+1=Sn-2n Sn+1Sn 两边同除以Sn+1 *Sn
得 1/Sn+1 - 1/Sn = 2n
以此类推 1/Sn - 1/Sn-1 = 2(n-1)
1/Sn-1 - 1/Sn-2 = 2(n-2)
...
1/S2 - 1/S1 = 2
以上各个式子相加得1/Sn - 1/S1 = 2(1+2+...+n)
=n(n+1)
而1/S1=1/2
=>Sn= 2n(n+1) / (n方+n+2)
再用Sn-Sn-1 可知 an
Sn+1=Sn-2n Sn+1Sn 两边同除以Sn+1 *Sn
得 1/Sn+1 - 1/Sn = 2n
以此类推 1/Sn - 1/Sn-1 = 2(n-1)
1/Sn-1 - 1/Sn-2 = 2(n-2)
...
1/S2 - 1/S1 = 2
以上各个式子相加得1/Sn - 1/S1 = 2(1+2+...+n)
=n(n+1)
而1/S1=1/2
=>Sn= 2n(n+1) / (n方+n+2)
再用Sn-Sn-1 可知 an
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