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已知函数f(x)=3x^2-2x,数列{an}的前n项和为Sn,点(n.Sn)在函数f(x)的图像上,数列{bn}满足
bn=3/anan+1,数列{bn}的前n项和为Tn,对于任意的n属于正整数,Tn
人气:346 ℃ 时间:2019-09-24 06:14:12
解答
点(n.Sn)在函数f(x)=3x^2-2x的图像上
Sn=3n^2-2n
S(n-1)=3(n-1)^2-2(n-1)
an=sn-s(n-1)=3(2n-1)-2=6n-5
设bn=3/(ana(n+1))
bn=3/(6n-5)(6n+1)=(1/2)[(1/(6n-5)-1/(6n+1)]
Tn=(1-1/7)/2+(1/7-1/13)/2+...+(1/(6n-5)-1/(6n+1)/2=(1-1/(6n+1)/2=3n/(6n+1)
3n/(6n+1)=(1-1/(6n+1)/2
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