Question

在数列{an}中，a1=1,a(n+1)=(1+a/n)an+(n+1)/2^n,设bn=an/n,求{bn}的通项公式，求{an}的前n项和

详细步骤

Answer 1

a(n+1)=(1+1/n)an+(n+1)/2^n=(n+1)(an/n+1/2^n),a(n+1)/(n+1)-an/n=1/2^n,an/n-a(n-1)/(n-1)=1/2^(n-1)┄┄a2/2-a1/1=1/2，上式两边相加得：a(n+1)/(n+1)--a1/1=1/2^n+1/2^(n-1)┄┄1/2=
1-1/2^n，a(n+1)/(n+1)=2-1/2^n,{bn}的通项公式=2-1/2^(n-1),
an=2n-n/2^(n-1),{an}的前n项和Sn=a1+a2+┄+an=2*(1+2+┄n)-[1/2^(1-1)+2/2^(2-1)+┄┄n/2^(n-1)]
1/2Sn=(1+2+┄n)-[1/2^(2-1)+2/2^(3-1)+┄┄+(n-1)/2^(n-1)+n/2^n],以上等式两边相减得：
1/2Sn=(1+2+┄n)-[1/2^(1-1)+1/2^(2-1)+1/2^(3-1)+┄┄+1/2^(n-1)-n/2^n],
=n(n+1)/2-[2-2/2^n-n/2^n]=n(n+1)/2+(n+2)/2^n-2
Sn=n(n+1)+(n+2)/2^(n-1)-4

Answer 2

很简单的

Answer 3

题目是否有问题(1+a/n)an表示什么意思