Sn为数列{an}的前n项和.a1=1,Sn=nan-2n(n-1)
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Sn为数列{an}的前n项和.a1=1,Sn=nan-2n(n-1)
(1)求证数列{an}为等差数列(2)设数列{1\anan1}的前n项和为Tn,求Tn
(1)求证数列{an}为等差数列(2)设数列{1\anan1}的前n项和为Tn,求Tn
![Sn为数列{an}的前n项和.a1=1,Sn=nan-2n(n-1)](/uploads/image/z/16401654-54-4.jpg?t=Sn%E4%B8%BA%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C.a1%3D1%2CSn%3Dnan-2n%28n-1%29)
解
当n>=2时
sn=nan-2n(n-1)=nan-2n²+2n
s(n-1)=(n-1)a(n-1)-2(n-1)(n-2)
=(n-1)a(n-1)-2n²+6n-4
∴
an=nan-(n-1)a(n-1)-4n+4
(n-1)a(n-1)=(n-1)an-4(n-1)
两边除以(n-1)
∴an-a(n-1)=4
又a1=1
∴an是以a1=1为首项,d=4为公差的等差数列
∴an=1+4(n-1)=4n-3
1/ana(n+1)=1/(4n-3)(4n+1)=1/4[1/(4n-3)-1/(4n+1)]
TN=1/4[(1-1/5)+(1/5-1/9)+……+(1/(4n-3)-1/(4n+1))]
=1/4[1-1/(4n+1)]
=1/4[4n/(4n+1)]
=n/(4n+1)
当n>=2时
sn=nan-2n(n-1)=nan-2n²+2n
s(n-1)=(n-1)a(n-1)-2(n-1)(n-2)
=(n-1)a(n-1)-2n²+6n-4
∴
an=nan-(n-1)a(n-1)-4n+4
(n-1)a(n-1)=(n-1)an-4(n-1)
两边除以(n-1)
∴an-a(n-1)=4
又a1=1
∴an是以a1=1为首项,d=4为公差的等差数列
∴an=1+4(n-1)=4n-3
1/ana(n+1)=1/(4n-3)(4n+1)=1/4[1/(4n-3)-1/(4n+1)]
TN=1/4[(1-1/5)+(1/5-1/9)+……+(1/(4n-3)-1/(4n+1))]
=1/4[1-1/(4n+1)]
=1/4[4n/(4n+1)]
=n/(4n+1)
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