Question

高中的数学题

Answer 1

(2n+1)an=(2n-1)a(n+1)
a(n+1)/an=(2n+1)/(2n-1)
累乘得:
(a2/a1)*(a3/a2)*...*a(n+1)/an=3*(5/3)*(7/5)*(9/7)*...*(2n+1)/(2n-1)
a(n+1)/a1=(2n+1)
a(n+1)=a1*(2n+1)
a(n+1)=2n+1
于是有:an=2n-1
a(n+1)-an=(2n+1)-(2n-1)=2
所以a(n+1)-an=2
即有{an}是公差为2,等差数列
2)bn=1/an*a(n+1)=1/[(2n+1)(2n-1)]=1/2[1/(2n-1)-1/(2n+1)]
Tn=1/2[(1-1/3)+(1/3-(1/5)+...+1/(2n-1)-1/(2n+1)]
Tn=1/2[1-1/(2n+1)]
Tn=n/(2n+1)

Answer 2

1、
即a(n+1)/an=(2n+1)/(2n-1)
所以
an/a(n-1)=(2n-1)/(2n-3)
a(n-1)/a(n-2)=(2n-3)/(2n-5)
……
a3/a2=5/3
a2/a1=3/1
相乘，中间约分
an/a1=(2n-1)/1
a1=1
所以an=2n-1
所以an-a(n-1)=2n-1-[2(n-1)-1]=2
所以an是等差数列

2、
bn=1/[(2n-1)(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
所以Tn=1/2*[(1-1/3)+(1/3-1/5)+……+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)

Answer 3

(2n+1)an =(2n-1)a(n+1)
a1=1
(1)
(2n+1)an =(2n-1)a(n+1)
a(n+1)/an = (2n+1)/(2n-1)
an/a(n-1)= (2n-1)/(2n-3)
an/a1 = [(2n-1)/(2n-3)].[(2n-3)/(2n-4)].....(3/1)
= 2n-1
an = 2n-1
=>{an} 是等差数列， d=2， a1=1
(2)
bn= 1/[an.a(n+1)]
=1/[(2n-1)(2n+1)]
=(1/2)[ 1/(2n-1)-1/(2n+1) ]
Tn =b1+b2+...+bn
=(1/2)[ 1 - 1/(2n+1) ]
= n/(2n+1)