Question

若数列{an}满足a1=2,a(n+1)=an+n-1,求数列{an}的通项公式

1.若数列{an}满足a1=2,a(n+1)=an+n-1,求数列{an}的通项公式
2.若数列{an}满足a1=1,a(n+1)=an*3^(n-1)
3.已知数列{an}满足a1=1,a(n+1)=sn+(n+1),n属于N*，求an和 sn

Answer 1

咦，原来还有两个问题啊，补下补下..
1.a(n+1)-an=n+1
a2-a1=2
a3-a2=3
.....
an-a(n-1)=n
an=(an-a(n-1))+(a(n-1)-a(n-2))+...+(a2-a1)+a1 (n>=2)
=n+(n-1)+...+3+2+2=n+(n-1)+...+3+2+1+1=(1+n)n/2+1=(n^2+n+2)/2
a1=2满足上式
an=(n^2+n+2)/2
2 a(n+1)/an=3^(n-1)
a2/a1=1
a3/a2=3
a4/a3=9
....
an/a(n-1)=3^(n-2)
an=an/a(n-1)*a(n-1)/a(n-2)*...*a2/a1 *a1=3^(n-2)*3^(n-3)*...*9*3*1*1=1*（1-3^(n-1）)/(1-3)
=(3^(n-1)-1)/2 (n>=2)
a1=1
3. a(n+1)=sn+(n+1), ....(1)
an=s(n-1)+n,...(2)
(1)-(2 ) a(n+1)-an=an+1
a(n+1)-2an=1
an-2a(n-1)=1=a(n+1)-2an
(a(n+1)-2an)/(an-2a(n-1)) =1 （n>=2）
(a3-2a2)/(a2-2a1)=1
(a4-2a3)/(a3-2a2)=1
.....
(an-2(an-1))/(an-1 -2(an-2))=1
依造之前两题思路整理下即可