Question

求数列{1/(2n-1)(2n+1)}的前n项和

Answer 1

原通项式=(1/2)*{1/(2n-1)-1/(2n+1)},求和时把各项的二分之一提到外面,相临两项抵消,剩第一项和最后一项

Answer 2

题目应该是数列{1/[(2n-1)(2n+1)]}的前n项和吧
Sn=1/(1*3)+1/(3*5)+...+1/[(2n-1)(2n+1)]
=1/2[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)

Answer 3

根据1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
可解

Answer 4

拆分
1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]

Sn=1/(1*3)-1/(3*5)+....+1/[(2n-1)(2n+1)]
=1/2[1/1-1/3+1/3-1/5+1/5....+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)]
前后消项得
=1/2[1-1/(2n+1)]
=1/2*2n/(2n+1)
=n/(2n+1)