(1)Sn=n(an-a1)/2,将n=1 代入
则S1=1(a1-a1)/2=0
又S1=a1 ,所以a1=0
故a=0;lz是对的哦!
(2)Sn=n(an-a1)/2=n*an/2
S(n-1)=(n-1)*a(n-1)/2
作差
Sn-S(n-1)=n*an/2-(n-1)a(n-1)/2
因为Sn-S(n-1)=an
所以an=n*an/2-(n-1)a(n-1)/2
通分并移项(n-1)a(n-1)=(n-2)an
an/a(n-1)=(n-1)/(n-2)
所以得到an=k(n-1),an 是等差数列
现在求系数k
当n=1时,a1=0,满足;
当n=2时,a2=k=p
故数列{an}的通项为an=p(n-1),是首项为0,公差为p的等差数列
(3)由题意:
bn=S(n+2)/S(n+1)+S(n+1)/S(n+2),
S(n+1)=[a1+a(n+1)](n+1)/2=n(n+1)p/2,
S(n+2)=[a1+a(n+2)](n+2)/2=(n+1)(n+2)p/2代入上式
得:bn=n/(n+2)+(n+2)/n=2+2[1/n-1/(n+2)]
故Tn=b1+b2+...+bn
=2+2(1-1/3)+2+2(1/2-1/4)+...+2+2[1/n-1/(n+2)]
=2n+2[1-1/3+1/2-1/4+...+1/n-1/(n+2)]
故:Tn-2n=2[1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=2{1/2[1+1/2-1/(n+1)-1/(n+2)}
=3-1/(n+1)-1/(n+2)
显然3-1/(n+1)-1/(n+2)<3
从而:Tn-2n<3得证!