Question

跪求大神解题高中数学，要过程

Answer 1

（1）S1=a1=0
(2)Sn-Sn-1=[n*an-(n-1)*an-1}=an
即an/an-1=(n-1)/(n-2) 所以an=n-1
a1=0 a2=1 均符合an
所以 {an} 为等差数列
（3）Sn=n(n-1)/2
bn=S(n+2)/S(n+1)+S(n+1)/S(n+2),
=(n+2)(n+1)/[(n+1)n]+(n+1)n/[(n+1)(n+2)]
= (n+2)/n+n/(n+2)
=1+2/n+1-2/(n+2)
=2+2/n-2/(n+2)
Tn=2n+[(2-2/3)+(1-2/4)+(2/3-2/5)+(2/4-2/6)+....+(2/(n-1)-2/(n+1))+(2/n-2/(n+2))
=2n+3-[2/(n+1)+2/(n+2)]
∵2/(n+1)+2/(n+2)>0 ∴ Tn<2n+3 对任意的n∈N*恒成立
∵ Tn-pn<q 等价于 Tn<pn+q
∴符合条件的实数p≥2,q≥3

Answer 2

这个题能倒过来么？看不清。。

追问

现在呢

会不会啊