问题
解答题
已知公差不为0的等差数列{an}的首项为4,设数列的前n项和为Sn,且
(1)求数列{an}的通项公式an及Sn; (2)记An=
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答案
(1)设等差数列{an}的公差为d,由(
)2=1 a2
•1 a1
,1 a4
得(a1+d)2=a1(a1+3d),因为d≠0,所以d=a1=4,
所以an=4n,Sn=4n+
=2n(n+1);4n(n-1) 2
(2)∵
=1 Sn
(1 2
-1 n
)1 n+1
∴An=
+1 S1
+…+1 S2
=1 Sn
(1-1 2
+1 2
-1 2
+…+1 3
-1 n
)=1 n+1
(1-1 2
).1 n+1
又a2n-1=4•2n-1=2n+1,
∴Bn=
+1 a1
+…+1 a2
=1 a2n-1
•1 4
=1-(
)n1 2 1- 1 2
(1-1 2
).1 2n
当n≥2时,2n=
+C 0n
+…+C 1n
>n+1,C nn
即1-
<1-1 n+1
.1 2n
所以An<Bn.