求教数学学霸，高中必修5，数列问题，谢谢
\u6c42\u6559\u6570\u5b66\u5b66\u9738\uff0c\u9ad8\u4e2d\u5fc5\u4fee5\uff0c\u6570\u5217\u95ee\u9898\uff0c\u8c22\u8c22Sn =a1+a2+...+an
Sn= n^2/(n.a+b)
n=1
a1= 1/(a+b)
1/2 =1/(a+b)
a+b=2 (1)
n=2
a1+a2 = 4/(2a+b)
1/2 +5/6 =4/(2a+b)
2a+b=3 (2)
(2)-(1)
a=1
from (1)
b=1
Sn= n^2/(n.a+b)
=n^2/(n+1)
an = Sn -S(n-1)
=n^2/(n+1) - (n-1)^2/n
= [n^3- (n+1)(n-1)^2]/ [n(n+1)]
= (n^2+n-1)/[n(n+1)]
a1= 1/2 satisfy an =(n^2+n-1)/[n(n+1)]
ie
an =(n^2+n-1)/[n(n+1)]
\u53d6n=17,S17/T17=17a9/17b9=a9/b9=88/33=8/3
11\u9898\u6709\u5468\u671f\u6027\uff0c
1/2,2,-1,1/2\u2026,
a2013=a3=-1
对第二式处理。分子第一项=2n-3，第二项=2n-1，变换成差的形式=（-1/（2n-1）+1/（2n-3））
列出bn，可以看出，sn=-1-1/（2n-1）=（-2n）/（2n-1）
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