已知数列{an}中,a1=1/2,点(n,2a(n+1)-an)(n∈N*)在直线y=x上.(1)计算a2,a3,a4的值;(2)令bn=a(n+1)-an-1,求证:数列{bn}是等比数列;(3)求数列{an}的通项公式.
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![已知数列{an}中,a1=1/2,点(n,2a(n+1)-an)(n∈N*)在直线y=x上.(1)计算a2,a3,a4的值;(2)令bn=a(n+1)-an-1,求证:数列{bn}是等比数列;(3)求数列{an}的通项公式.](/uploads/image/z/3977328-48-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2F2%2C%E7%82%B9%28n%2C2a%28n%2B1%29-an%29%28n%E2%88%88N%2A%29%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dx%E4%B8%8A.%281%29%E8%AE%A1%E7%AE%97a2%2Ca3%2Ca4%E7%9A%84%E5%80%BC%3B%282%29%E4%BB%A4bn%3Da%28n%2B1%29-an-1%2C%E6%B1%82%E8%AF%81%3A%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%3B%283%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.)
已知数列{an}中,a1=1/2,点(n,2a(n+1)-an)(n∈N*)在直线y=x上.(1)计算a2,a3,a4的值;(2)令bn=a(n+1)-an-1,求证:数列{bn}是等比数列;(3)求数列{an}的通项公式.
已知数列{an}中,a1=1/2,点(n,2a(n+1)-an)(n∈N*)在直线y=x上.
(1)计算a2,a3,a4的值;
(2)令bn=a(n+1)-an-1,求证:数列{bn}是等比数列;
(3)求数列{an}的通项公式.
已知数列{an}中,a1=1/2,点(n,2a(n+1)-an)(n∈N*)在直线y=x上.(1)计算a2,a3,a4的值;(2)令bn=a(n+1)-an-1,求证:数列{bn}是等比数列;(3)求数列{an}的通项公式.
(1)由题意得n=2a(n+1)-an则a(n+1)=(n+an)/2 下面的题还需用到n-a(n+1)=a(n+1)-an
a2=3/4
a3=11/8
a4=35/16
(2)bn=a(n+1)-an-1=n-a(n+1)-1
b(n+1)=a(n+2)-a(n+1)-1=(n+1+a(n+1))/2-a(n+1)-1=(n-a(n+1)-1)/2
b(n+1)/bn=1/2常数
所以,数列{bn}是等比数列
(3)b1=a2-a1-1=-3/4
bn=-3/4*(1/2)^(n-1)
a(n+1)-an=bn+1=1-3/4*(1/2)^(n-1)
an-a(n-1)=b(n-1)+1=1-3/4*(1/2)^(n-2)
.
a3-a2=1-3/8
a2-a1=1-3/4
以上各式求和则有
an-a1=n-1-3/2[1-(1/2)^(n-1)]
an=n-2+3/2*(1/2)^(n-1)