问题
解答题
已知数列{an}中,a1=
(1)令bn=an+1-an-1,求证数列{bn}是等比数列; (2)求数列{an}的通项. |
答案
(1)证明:a1=
,2an+1=an+n,1 2
∵a2=
,a2-a1-1=3 4
-3 4
-1=-1 2
,3 4
又bn=an+1-an-1,bn+1=an+2-an+1-1,
∴
=bn+1 bn an+2-an+1-1 an+1-an-1
=
=
-an+1+(n+1) 2
-1an+n 2 an+1-an-1
=an+1-an-1 2 an+1-an-1
.1 2
bn=-
×(3 4
)n-1=-1 2
×3 2
,1 2n
∴{bn}是以-
为首项,以3 4
为公比的等比数列.1 2
(2)∵an+1-an-1=-
×3 2
,1 2n
∴a2-a1-1=-
×3 2
,1 2
a3-a2-1=-
×3 2
,1 22
∴an-an-1-1=-
×3 2
,1 2n^-1
将以上各式相加得:
∴an-a1-(n-1)=-
(3 2
+1 2
++1 22
),1 2n^-1
∴an=a1+n-1-
×3 2
(1-1 2
)1 2n-1 1- 1 2
=
+(n-1)-1 2
(1-3 2
)=1 2n-1
+n-2.3 2n
∴an=
+n-2.3 2n