在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 15:43:28
![在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;](/uploads/image/z/2813293-37-3.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D2%2Can%3D2an-1%2B2%5E%28n%2B1%29%28n%3E%3D2%2C%29%E4%BB%A4bn%3Dan%2F2%5En%2C%E6%B1%82%E8%AF%81bn%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E5%86%99%E5%87%BA%E5%85%B6%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%3B)
在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;
在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;
在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;
将an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n两边除以2^n,得an/2^n=2a(n-1)/2^n+2,即
bn=a(n-1)/2^(n-1)+2,所以bn=b(n-1)+2,所以bn是等差数列.b1=a1/2=1,所以bn=2n-1
∵an=2a(n-1)+2^(n+1)
且2^n≠0,等式两边同乘1/2^n,可得:
an/2^n = [2a(n-1)]/(2^n) + 2^(n+1)/2n
an/2^n - [a(n-1)]/[2^(n-1)] = 2
∵bn=an/(2^n),带入上式,可得:
bn - b(n-1) = 2
∴数列...
全部展开
∵an=2a(n-1)+2^(n+1)
且2^n≠0,等式两边同乘1/2^n,可得:
an/2^n = [2a(n-1)]/(2^n) + 2^(n+1)/2n
an/2^n - [a(n-1)]/[2^(n-1)] = 2
∵bn=an/(2^n),带入上式,可得:
bn - b(n-1) = 2
∴数列{bn}是以b1=a1/2=1为首项,公差为2的等差数列
∴bn=b1+(n-1)d=1+2(n-1) = 2n-1
另:∵an=bn (2^n)
=(2n-1)(2^n)
也求得了an
收起
an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n
bn=an-1/2^(n-1)+2,则bn+1=an/2^n+2
bn+1-bn=an/2^n-an-1/2^(n-1)=an-2an-1/2^n=2^(n+1)/2^n=2
所以bn是公差为2的等差数列
由bn=an/2^n 可知
b(n-1)=a(n-1)/2^(n-1)
bn-b(n-1)=an/2^n-a(n-1)/2^(n-1)=[an-2a(n-1)]/(2^n)
由an=2an-1+2^(n+1)(n>=2,) 可知 an-2a(n-1)=2^(n+1)
所以[an-2a(n-1)]/(2^n)=2 所以bn-b(n-1)=2
所以bn是等...
全部展开
由bn=an/2^n 可知
b(n-1)=a(n-1)/2^(n-1)
bn-b(n-1)=an/2^n-a(n-1)/2^(n-1)=[an-2a(n-1)]/(2^n)
由an=2an-1+2^(n+1)(n>=2,) 可知 an-2a(n-1)=2^(n+1)
所以[an-2a(n-1)]/(2^n)=2 所以bn-b(n-1)=2
所以bn是等差数列 公差为2
b1=a1/2^1 且a1=2 所以 b1=1
所以bn的通项公式为 bn=1+2(n-1) (n>1)
收起