已知首项为负的数列{an}中,相邻两项不为相反数,且前n项和Sn=1/4(an-5)(an+7)
1个回答
1)an=Sn-S_(n-1)
an =1/4(an-5)(an+7)-1/4(a_(n-1)-5)(a_(n-1)+7)
4an=an^2+2an-(a_n-1)^2-2(a_n-1)
(an+a_n-1)(an-a_n-1)-2(an+a_n-1)=0
又相邻两项不为相反数 则an+a_n-1不等于0
可得an-a_n-1=2
问题得证
2）n=1时S1=a1=1/4(a1-5)(a1+7)a1为负可得a1=-5
Tn=1/a1a2+1/a2a3+..+1/anan+1
又1/a1-1/a2=(a2-a1)/a1a2=2/a1a2
同理可得Tn=1/2*{1/a1-1/a2+1/a2-1/a3+...+1/an-1/an+1}
=1/2*(1/a1-1/an+1)
=1/2*(an+1-a1)/a1an+1
=1/2*(2n)/a1(a1+2n)
=n/(a1^2+2na1)=n/(25-10n)
1/M>= 1/Tn=25/n-10>=-10 则M
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