Question

17题，在线等，急。

Answer 1

(1)
n≥2时，
2an=2Sn-2S(n-1)=(n+1)an+1-[na(n-1)+1]
(n-1)an=na(n-1)
an/n=a(n-1)/(n-1)
a1/1=(3/2)/1=3/2
数列{an/n}是各项均为3/2的常数数列
an/n=3/2
an=3n/2
n=1时，a1=3/2，同样满足表达式
数列{an}的通项公式为an=3n/2
(2)
bn=1/(an+1)²
=1/(3n/2 +1)²
=4/(3n+2)²
<4/[(3n-1)(3n+2)]
=(4/3)[1/(3n-1)- 1/(3(n+1)-1)]
Tn=b1+b2+...+bn
<(4/3)[1/2-1/5 +1/5 -1/8+...+1/(3n-1) -1/(3n+2)]
=(4/3)[½ -1/(3n+2)]
1/(3n+2)>0，½ -1/(3n+2)<½
(4/3)[½- 1/(3n+2)]<(4/3)·½=⅔<7/10
Tn<7/10