设数列{an}的前n项和为Sn,点（n,Sn/n)(n属于N正）均在函数y=3x-2的图象上设bn=3/AnA(n+1),Tn是数列{bn}的前n项和,求Tn

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设数列{an}的前n项和为Sn,点（n,Sn/n)(n属于N正）均在函数y=3x-2的图象上设bn=3/AnA(n+1),Tn是数列{bn}的前n项和,求Tn
设数列{an}的前n项和为Sn,点（n,Sn/n)(n属于N正）均在函数y=3x-2的图象上
设bn=3/AnA(n+1),Tn是数列{bn}的前n项和,
求Tn

设数列{an}的前n项和为Sn,点（n,Sn/n)(n属于N正）均在函数y=3x-2的图象上设bn=3/AnA(n+1),Tn是数列{bn}的前n项和,求Tn
∵点(n,Sn/n)(n属于N正)均在函数y=3x-2的图象上
∴Sn/n = 3n-2 ,即：Sn=3n^2 - 2n
则：S(n-1)=3(n-1)^2 - 2(n-1) =3n^2 - 8n + 5
两式相减,得：Sn - S(n-1)=6n-5
即：an=6n-5
则a(n+1)=6(n+1)-5=6n+1
bn=3/AnA(n+1) =3/(6n-5)(6n+1)
=3*(1/6)*[1/(6n-5) - 1/(6n+1)]
=(1/2)*[1/(6n-5) - 1/(6n+1)]
则有b1=(1/2)*(1/1 - 1/7)
b2=(1/2)*(1/7 - 1/13)
…
…
bn=(1/2)*[1/(6n-5)-1/(6n+1)]
∴Tn=b1+b2+b3+.+bn
=(1/2)*{(1/1 - 1/7)+(1/7 - 1/13)+.+[1/(6n-5) - 1/(6n+1)]}
=(1/2)*[1-1/(6n+1)]
=3n/(6n+1)

点（n,Sn/n)(n属于N正）看不明白
3/AnA(n+1), 什么意思？
最好描述一下

步骤1：求出{An}是等差数列，An=6n-5
步骤2：Tn=B1+B2+……Bn用裂项相消法
结果：（1/2）*[1-1/(6n+1)]

全部展开

因为点（n,Sn/n)(n属于N正）均在函数y=3x-2的图象上
所以Sn/n=3n-2
那么Sn=n(3n-2)=3n^2-2n
故n≥2时
an=Sn-S(n-1)=3n^2-2n-[3(n-1)^2-2(n-1)]=6n-5
n=1时
a1=S1=3*1^2-2*1=1也满足通项an=6n-5
所以an=6n-5
bn=3/ana(n+1)=3/(6n-5)(6n+1)=[1/(6n-5)-1/(6n+1)]/2
所以{bn}的前n项和是Tn=b1+b2+...+bn
=[(1-1/7)+(1/7-1/13)+...+(1/(6n-5)-1/(6n+1))]/2
=[1-1/(6n+1)]/2
=3n/(6n+1)

收起

先把点代入直线方程，s（n+1）减sn算出an，2bn=1/An-1/A(n+1),裂项相消即可