在数列an中,a2 1 4,且(n-an)an+1(n

2021-02-24 18:15:35 字数 2395 阅读 5592

1楼:匿名用户

^(n-an)a(n+1)=(n-1)an

a(n+1) = (n-1)an/ (n- an)

1/a(n+1) = (n-an)/[(n-1)an]

= n/[(n-1)an] - 1/(n-1)

1/[na(n+1)] - 1/[(n-1)an] = -1/n(n-1)

= [ 1/n -1/(n-1) ]

1/[(n-1)an] - 1/[(n-2)a(n-1)] = [ 1/(n-1) - 1/(n-2) ]

1/[(n-1)an] - 1/a2 = 1/(n-1) - 1

1/[(n-1)an] = 1/(n-1) +3

= (3n-2)/(n-1)

an = 1/(3n-2)

ie an = 1/(3n-2)

by mi

an =1/(3n-2)

(1-a1)a2=0

a1=1

p(1) is true

assume p(k) is true

ieak = 1/(3k-2)

for n=k+1

(k-ak)a(k+1)=(k-1)ak

a(k+1) =(k-1)ak/(k-ak)

=[(k-1)/(3k-2)]/[k-1/(3k-2)]

= (k-1)/(3k^e68a84e8a2ad62616964757a686964616f313333353330332-2k-1)]

= (k-1)/(3k+1)(k-1)]

=1/(3k+1)

p(k+1) is true

by principle of mi, it is true for all +ve integer n

a1=1

a3=1/(9-2)=1/7

a4=1/(12-2)=1/10

已知数列{an}中,an=1+1/[a+2(n-1)],(n∈n+,a∈r,且a≠0)。

2楼:庐山醉仙

答:抄(1)当a=-7时,an=1+1/[a+2(n-1)]=1+1/[-7+2(n-1)]=1+1/2n-9。an是个减函数,当n=1时,an取最大值:

6/7。当n趋近于无穷大时,an的极限值就是:1。

即,最大值是6/7,极小值1。.

(2)由an=1+1/[a+2(n-1)],an≤a6,可得:1+1/[a+2(n-1)]≤1+1/[a+2(6-1)],即:n>=6。

在数列{an}中,已知a1=1,且满足an+1-an=an/(n+1),求通项公式.

3楼:钟馗降魔剑

∵a(n+1)-an=an/(n+1)

∴a(n+1)=an+an/(n+1)

=an*(n+2)/(n+1)

∴a(n+1)/an=(n+2)/(n+1)那么an/a(n-1)=(n+1)/n

a(n-1)/a(n-2)=n/(n-1)…………………………

a3/a2=4/3

a2/a1=3/2

累乘,得:an/a1=(n+1)/2

而a1=1,∴an=(n+1)/2

4楼:我不是他舅

移项a(n+1)=(n+2)/(n+1)*ana(n+1)/an=(n+2)/(n+1)所以an/a(n-1)=(n+1)/n

……a3/a2=4/3

a2/a1=3/2

相乘an/a1=(n+1)/n

所以an=(n+1)/n

5楼:不知道后才知道

通分,求得an=n+1╱n因为

数列{an}中,a1=2,a2=1,2/an=1/an+1+1/an-1(n>1)则an=?

6楼:匿名用户

根据我的理解,数列是首项为1/a1=1/2 公差d=1/a2-1/a1=1-1/2=1/2

的等差数列 ∴有1/an=1/2+(n-1)*1/2=n/2

所以an的通项an=2/n

7楼:梦雅盈月

数列是首项

为1/a1=1/2 公差d=1/a2-1/a1=1-1/2=1/2的等差数列 ∴有1/an=1/2+(n-1)*1/2=n/2所以an的通项an=2/n

得1/an-1/a(n-1)=1/2

1/a2-1/a1=1/2

8楼:匿名用户

由原式得1/a(n+1)-1/an=1/an-1/a(n-1),对任何an都成立,则其必等于常数,将前二个数代入得1/an-1/a(n-1)=1/2

1/a2-1/a1=1/2

1/an-1/a(n-1)=1/2

直到n累和可求an

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