设数列{an}的首项a1=1,且点(an,an+1)在直线x+2y=0上.

题目

设数列{an}的首项a1=1,且点(an,an+1)在直线x+2y=0上.
1,求数列{an}的前n项和sn.2,若数列{an}满足bn=na1+(n-1)a2+…+2an-1+an,求bn通项公式.

答案

依题意
an+2a(n+1)=0;
a(n+1)=-an/2
an=a1*q^(n-1)=1*(-1/2)^(n-1)=(-1/2)^(n-1);
Sn=[1-(-1/2)^n]/[1-(-1/2)]=(2/3)*[1-（-1/2)^n)]
2
b(n+1)-bn=a1+a2+```+an=Sn;
bn-b(n-1)=S(n-1)
```
b2-b1=S1
bn=(bn-b(n-1))+(b(n-1)-b(n-2))+```+b2-b1+b1
=S(n-1)+S(n-2)+```+S1+b1
再求和即可