已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2•a3=45,a1+a4=14.(1)求数列{an}的通项公式;(2)令bn=2Sn2n-1,f(n)=bn(n+25)•bn+1(n

题目简介

已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2•a3=45,a1+a4=14.(1)求数列{an}的通项公式;(2)令bn=2Sn2n-1,f(n)=bn(n+25)•bn+1(n

题目详情

已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2•a3=45,a1+a4=14.
(1)求数列{an}的通项公式;
(2)令bn=
2Sn
2n-1
,f(n)=
bn
(n+25)•bn+1
(n∈N*),求f(n)的最大值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵数列an}是等差数列,
∴a2•a3=45,a1+a4=a2+a3=14.
a2=3
a3=9
a2=9
a3=5

∵公差d>0,
a2=3
a3=9
,解得d=4,a1=1.
∴an=1+4(n-1)=4n-3.
(Ⅱ)∵Sn=na1+
n(n-1)d
2
=2n2-n

bn=
2Sn
2n-1
=2n,
∴f(n)=
bn
(n+25)•bn+1
=class="stub"2n
(n+25)⋅2(n+1)
=class="stub"n
n2+26n+5
=class="stub"1
n+class="stub"25
n
+26
≤class="stub"1
26+2
class="stub"25
n
•n
=class="stub"1
26+10
=class="stub"1
36

当且仅当n=class="stub"25
n
,即n=5时,f(n)取得最大值class="stub"1
36

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