问题
解答题
已知数列{an}的各项均是正数,其前n项和为Sn,满足Sn=4-an. (1)求数列{an}的通项公式; (2)设bn=
|
答案
(1)由Sn=4-an.得S1=4-a1,解得a1=2,
而an+1=Sn+1-Sn=(4-an+1)-(4-an)=an-an+1,即2an+1=an,
∴
=an+1 an
,1 2
可见,数列{an}是首项为2,公比为
的等比数列.1 2
∴an=2•(
)n-1=(1 2
)n-2;1 2
(2)证明:∵bn=
=1 2-log2an
=1 2-(2-n)
,1 n
∴bnbn+2=
=1 n(n+2)
(1 2
-1 n
),1 n+2
∴数列{bnbn+2}的前n项和
Tn=
[(1-1 2
)+(1 3
-1 2
)+(1 4
-1 3
)+(1 5
-1 4
)+…+(1 6
-1 n-2
)+(1 n
-1 n-1
)+(1 n+1
-1 n
)]1 n+2
=
(1+1 2
-1 2
-1 n+1
)1 n+2
=
(1 2
-3 2
-1 n+1
)=1 n+2
-3 4
(1 2
+1 n+1
)<1 n+2
.3 4