已知数列{an}满足a1=4,an=4-4/an-1已知数列{an}满足a1=4,an=4-4/a(n-1)(n≥2),令bn=1/ an-2.1、求证:数列{bn}是等差数列 2、求数列{an}通项
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![已知数列{an}满足a1=4,an=4-4/an-1已知数列{an}满足a1=4,an=4-4/a(n-1)(n≥2),令bn=1/ an-2.1、求证:数列{bn}是等差数列 2、求数列{an}通项](/uploads/image/z/12706920-0-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D4%2Can%3D4-4%2Fan-1%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D4%2Can%3D4-4%2Fa%EF%BC%88n-1%EF%BC%89%28n%E2%89%A52%29%2C%E4%BB%A4bn%3D1%2F+an-2.1%E3%80%81%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+2%E3%80%81%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E9%80%9A%E9%A1%B9)
已知数列{an}满足a1=4,an=4-4/an-1已知数列{an}满足a1=4,an=4-4/a(n-1)(n≥2),令bn=1/ an-2.1、求证:数列{bn}是等差数列 2、求数列{an}通项
已知数列{an}满足a1=4,an=4-4/an-1
已知数列{an}满足a1=4,an=4-4/a(n-1)(n≥2),令bn=1/ an-2.1、求证:数列{bn}是等差数列 2、求数列{an}通项
已知数列{an}满足a1=4,an=4-4/an-1已知数列{an}满足a1=4,an=4-4/a(n-1)(n≥2),令bn=1/ an-2.1、求证:数列{bn}是等差数列 2、求数列{an}通项
an=4-4/a(n-1)
an-2
=2-4/a(n-1)
=2{[a(n-1)-2]/a(n-1)}
1/(an-2)=1/2+1/[a(n-1)-2]、
1/(a1-2)=1/2
bn=1/(an-2)=1/2+b(n-1)
bn-b(n-1)=1/2
所以数列{bn}是以1/2为首项,以1/2为公关的等差数列,
bn=n/2
1/(an-2)=n/2
得an=(2/n)+2
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(1)证明:
an=4-4/a(n-1)
a(n)-2=2- 4/a(n-1)
b(n)=1/(a(n)-2)=1/[2-4/a(n-1)]=a(n-1)/[2a(n-1)-4]=a(n-1)/2[a(n-1)-2]=1/2 +1/[a(n-1)-2]
bn-b(n-1)=[1/(an-2)]=1/2为定值
所以数列{bn}是等差数列
(2)
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(1)证明:
an=4-4/a(n-1)
a(n)-2=2- 4/a(n-1)
b(n)=1/(a(n)-2)=1/[2-4/a(n-1)]=a(n-1)/[2a(n-1)-4]=a(n-1)/2[a(n-1)-2]=1/2 +1/[a(n-1)-2]
bn-b(n-1)=[1/(an-2)]=1/2为定值
所以数列{bn}是等差数列
(2)
{bn}首项是1/(4-2)=1/2,公差是1/2
bn=1/2+(1/2)*(n-1)=n/2
1/(an-2)=n/2
an-2=2/n
an=2+2/n
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(1)
an=4-4/a(n-1)
an -2 = 2(a(n-1) - 2)/a(n-1)
1/[an -2 ] = a(n-1)/[ 2(a(n-1) - 2) ]
= 1/2 + 1/(a(n-1) - 2)
1/[an -2 ] -1/(a(n-1) - 2) = 1/2
bn =1/(an - 2)
{...
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(1)
an=4-4/a(n-1)
an -2 = 2(a(n-1) - 2)/a(n-1)
1/[an -2 ] = a(n-1)/[ 2(a(n-1) - 2) ]
= 1/2 + 1/(a(n-1) - 2)
1/[an -2 ] -1/(a(n-1) - 2) = 1/2
bn =1/(an - 2)
{bn}是等差数列, d=1/2
(2)
1/[an -2 ] -1/(a(n-1) - 2) = 1/2
1/[an -2 ] -1/(a1 - 2) = (n-1)/2
1/[an -2 ] =n/2
an = 2+ 2/n
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