2个回答
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解:方法一、
n=1时,a1=3=6/2
n=2时,a2=8/3
n=3时,a3=10/4
于是猜测an=2(n+2)/(n+1)
下面用数学归纳法来证明
证明:当n=1时, a1=2(1+2)/(1+1)=3,式子an=2(n+2)/(n+1)成立,
假设当n=k时,有ak=2(k+2)/(k+1)成立,
则当n=k+1时,a(k+1)=4-4/ak=4-2(k+1)/(k+2)=(2k+6)/(k+2)=2(k+1+2)/(k+1+1)也成立
所以对一切正整数n都有an=2(n+2)/(n+1)成立。
方法二、
因为a(n+1) = 4- 4/an
所以a(n+1)-2=2-4/an
即a(n+1) - 2 = 2(an-2)/an
即1/[a(n+1) - 2] = an/[2(an-2)]
即1/[a(n+1) - 2] = [an-2+2]/[2(an-2)]
即 1/[a(n+1) - 2]= 1/2 + 1/(an-2)
即1/[a(n+1) - 2] -1/(an-2) =1/2
于是数列{1/(an-2)}是以1/(a1-2)=1为首项,公差为1/2的等差数列
所以1/(an-2) =1+(1/2)(n-1)
解得an = 2(n+2)/(n+1)
n=1时,a1=3=6/2
n=2时,a2=8/3
n=3时,a3=10/4
于是猜测an=2(n+2)/(n+1)
下面用数学归纳法来证明
证明:当n=1时, a1=2(1+2)/(1+1)=3,式子an=2(n+2)/(n+1)成立,
假设当n=k时,有ak=2(k+2)/(k+1)成立,
则当n=k+1时,a(k+1)=4-4/ak=4-2(k+1)/(k+2)=(2k+6)/(k+2)=2(k+1+2)/(k+1+1)也成立
所以对一切正整数n都有an=2(n+2)/(n+1)成立。
方法二、
因为a(n+1) = 4- 4/an
所以a(n+1)-2=2-4/an
即a(n+1) - 2 = 2(an-2)/an
即1/[a(n+1) - 2] = an/[2(an-2)]
即1/[a(n+1) - 2] = [an-2+2]/[2(an-2)]
即 1/[a(n+1) - 2]= 1/2 + 1/(an-2)
即1/[a(n+1) - 2] -1/(an-2) =1/2
于是数列{1/(an-2)}是以1/(a1-2)=1为首项,公差为1/2的等差数列
所以1/(an-2) =1+(1/2)(n-1)
解得an = 2(n+2)/(n+1)
追问
也谢谢你了
展开全部
a(n+1) = 4- 4/an
= (4an-4)/an
a(n+1) - 2 = 2(an-2)/an
1/[a(n+1) - 2] = an/[2(an-2)]
= 1/2 + 1/(an-2)
1/[a(n+1) - 2] -1/(an-2) =1/2
{1/(an-2)}是等差数列, d=1/2
1/(an-2) - 1/(a1-2) = (n-1)/2
1/(an-2) = (n+1)/2
an-2 = 2/(n+1)
an = 2+ [ 2/(n+1) ]
= (4an-4)/an
a(n+1) - 2 = 2(an-2)/an
1/[a(n+1) - 2] = an/[2(an-2)]
= 1/2 + 1/(an-2)
1/[a(n+1) - 2] -1/(an-2) =1/2
{1/(an-2)}是等差数列, d=1/2
1/(an-2) - 1/(a1-2) = (n-1)/2
1/(an-2) = (n+1)/2
an-2 = 2/(n+1)
an = 2+ [ 2/(n+1) ]
追问
谢谢,但我想问下,解这种题目题目的思路一般是怎么样的?尤其是a(n+1)-2=2(an-2)/an这一步的想法是怎么得的???
追答
e.g
a(n+1) = (4an-4)/an
The aux. equation
x = (4x-4)/x
x^2-4x+4=0
(x-2)^2=0
x=2
a(n+1) = (4an-4)/an
a(n+1) -2= (4an-4)/an -2
...
....
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