mashang123456
幼苗
共回答了23个问题采纳率:91.3% 举报
1)依题意,
可设为:f(x)=a(x-1/3)^2-1/3,把原点(0,0)代入得:a=3,
于是代入化简得:f(x)=3x^2-2x
(2)依题意:Sn=3n^2-2n,有a1=S1=1,当n>=2时,an=Sn-Sn-1=6n-5.
而a1满足上式,所以an=6n-5,n属于正整数.
(3)根据(2),bn=1/an*an+1=1/(6n-5)*(6n+1)=(1/6)*[(1/6n-5)-(1/6n+1)]
于是:Tn=b1+b2+...+bn=(1/6)*[(1-1/7)+(1/7-1/13)+...+(1/6n-11)-(1/6n-5)+(1/6n-5)-(1/6n+1)]
=(1/6)*[1-(1/6n+1)=n/6n+1.
令Tn=(10/3)*(1-1/7)=20/7,
另一方面20n/6n+1=(10/3)*(1-1/6n+1)
1年前
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