在数列A(n)中,A(1)=1,A(n+1)=(1+1/n)An+(n+1)/2^n设Bn=An/n,求Bn的通项公式求数列An的前n项和Sn
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![在数列A(n)中,A(1)=1,A(n+1)=(1+1/n)An+(n+1)/2^n设Bn=An/n,求Bn的通项公式求数列An的前n项和Sn](/uploads/image/z/1159262-62-2.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97A%EF%BC%88n%EF%BC%89%E4%B8%AD%2CA%EF%BC%881%EF%BC%89%3D1%2CA%EF%BC%88n%2B1%EF%BC%89%3D%EF%BC%881%2B1%2Fn%EF%BC%89An%2B%EF%BC%88n%2B1%EF%BC%89%2F2%5En%E8%AE%BEBn%3DAn%2Fn%2C%E6%B1%82Bn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E6%B1%82%E6%95%B0%E5%88%97An%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
在数列A(n)中,A(1)=1,A(n+1)=(1+1/n)An+(n+1)/2^n设Bn=An/n,求Bn的通项公式求数列An的前n项和Sn
在数列A(n)中,A(1)=1,A(n+1)=(1+1/n)An+(n+1)/2^n
设Bn=An/n,求Bn的通项公式
求数列An的前n项和Sn
在数列A(n)中,A(1)=1,A(n+1)=(1+1/n)An+(n+1)/2^n设Bn=An/n,求Bn的通项公式求数列An的前n项和Sn
(1)B(1)=A(1)/1=1
由A(n+1)=(1+1/n)A(n)+(n+1)/2^n=(n+1)A(n)/n+(n+1)/2^n
A(n+1)/(n+1)=A(n)/n+1/2^n
即B(n+1)=B(n)+1/2^n
B(n)=B(n-1)+1/2^(n-1)
=B(n-2)+1/2^(n-2)+1/2^(n-1)
=…
=B(1)+1/2+…+1/2^(n-1)
=1+1/2+…+1/2^(n-1)
=2-1/2^(n-1);
(2)A(n)=nB(n)=2n-n/2^(n-1)
S(n)=2(1+2+…+n)-[1+2/2+3/4+…+n/2^(n-1)]=n^2+n-T(n)
其中T(n)=1+2/2+3/4+4/8+…+n/2^(n-1)
2T(n)=2+2/1+3/2+4/4+…+n/2^(n-2)
两式相减T(n)=2+1+1/2+1/4+…+1/2^(n-2)-n/2^(n-1)=4-(n+2)/2^(n-1)
S(n)=n^2+n-4+(n+2)/2^(n-1).
A(n+1)=An+An/n+1/2^n+n/2^n,∴(n+1)Bn+(n+1)/2^n=A(n+1)∴B(n+1)=Bn+1/2^n∴∑{B(n+1)-Bn}=∑1/2^n,故Bn=B1+∑1/2^(n-1),即Bn=2-1/2^(n-1)
An=2n-n/2^(n-1),然后错位相减
Sn=n^2+n-4+(n+2)/2^(n-1)