已知数列{an}满足a1=1,an=3n-1+an-1(n≥2). (Ⅰ)求a2,a3; (Ⅱ)证明an=3n−12.

问题描述:

已知数列{an}满足a1=1,an=3n-1+an-1(n≥2).
(Ⅰ)求a2,a3
(Ⅱ)证明an

3n−1
2

(Ⅰ)∵a1=1,
∴a2=3+1=4,
∴a3=32+4=13;
(Ⅱ)证明:由已知an-an-1=3n-1,n≥2
故an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=3n−1+3n−2+…+3+1=

3n−1
2
.n≥2
当n=1时,也满足上式.
所以an
3n−1
2