已知数列{an}中,a1=5/6,an+1=1/3an+(1/2)n+1,求an.

问题描述:

已知数列{an}中,a1=

5
6
,an+1=
1
3
an+(
1
2
n+1,求an

∵数列{an}中,a1=

5
6
,an+1=
1
3
an+(
1
2
n+1
两边同时乘以3n+1,得3n+1an+13nan+(
3
2
)n+1

从而3n+1an+1-3ⁿan=(
3
2
n+1
从而有:
3ⁿan-3n+1an-1=(
3
2
)ⁿ,
3n+1an-1-3n+2an-2=(
3
2
n+1
32a2-3a1=(
3
2
2
3a1=
5
2

累加得3ⁿan=3(
3
2
)ⁿ-2,
故an=
3
2n
-
2
3n

综上,数列{an}的通项公式为an=
3
2n
2
3n