Question

数学题 求解 谢谢

已知数列{An}的通项公式An=1/(2n_1)(2n+1),求前n项的和

Answer 1

An=[2/(2n-1)]-[2/(2n+1)]
所以
Sn=A1+A2+......+An=(2/1-2/3)+(2/3-2/5)+......+[2/(2n-1)]-[2/(2n+1)]=2-[2/(2n-1)]

Answer 2

• 通项=2/(2N-1)-2/(2N+1)

Answer 3

2an=2/(2n-1)(2n+1)
=[(2n+1)-(2n-1)]/(2n-1)(2n+1)
=(2n+1)/(2n-1)(2n+1)-(2n-1)/(2n-1)(2n+1)
=1/(2n-1)-1/(2n+1)
所以an=(1/2)[1/(2n-1)-1/(2n+1)]
Sn=1/2[1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)
=1/2[1-1/(2n+1)
=n/(2n+1)