Question

高中数学奥赛题

已知数列{a n}中，a n =[(2n+1)^2+1]/[(2n+1)^2-1]，求其前n项的和S n

（题目中空格后的字母均为下标）

Answer 1

LS过程有明显错误

解:
由于:
an
=[(2n+1)^2+1]/[(2n+1)^2-1]
=[(4n^2+4n+1)+1]/[(2n+1+1)(2n-1-1)]
=[4(n^2+n)+2]/[(2n+2)(2n)]
=[4n(n+1)+2]/[4(n+1)n]
=1+1/[2n(n+1)]
=1+(1/2)[1/(n+1)n]
则：
Sn
=a1+a2+...+an
={1+(1/2)[1/(2*1)]}+{1+(1/2)[1/(3*2)]}+...+
{1+(1/2)[1/(n+1)*n]}
=n+(1/2)[1/(1*2)+1/(2*3)+...+1/n*(n+1)]
=n+(1/2)[(1-1/2)+(1/2-1/3)+(1/n-1/(n+1))]
=n+(1/2)[1-1/(n+1)]
=n+(1/2)[n/(n+1)]
=[n(2n+3)]/(2n+2)

Answer 2

An=[(2n+1)^2+1]/[(2n+1-1)(2n+1+1)]
=[(2n+1)^2+1]/[2n*(2n+2)]
=(4n^2+4n+2)/[2n*(2n+2)]
=(2n^2+2n+1)/[n*(n+1)]
=[n^2+(n+1)^2]/[n*(n+1)]
=[n^2/(n*(n+1))]+[(n+1)^2/(n*(n+1))]
=n/(n+1)+(n+1)/n
=1
Sn=1+1+1+……+1=n

Answer 3

a n=[(2n+1)^2-1+2]/[(2n+1)^2-1]=1+2/[(2n+1)^2-1]
=1+2/〔(2n+2)2n〕
所以Sn=n+2/4*2+2/6*4+2/8*6………2/〔(2n+2)2n〕
=n+2*1/2*〔1/2-1/4+1/4-1/6+1/6-1/8+………1/2n-1/(2n+2)〕
=n+〔1/2-1/(2n+2)〕
=(2n^2+3n)/(2n+2)