题目

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

提问时间：2021-01-08

答案

∵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)