已知公差大于零的等差数列{an},前n项和为Sn.且满足a3a4=117,a2+a5=22.
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已知公差大于零的等差数列{an},前n项和为Sn.且满足a3a4=117,a2+a5=22.
(Ⅰ)求数列an的通项公式;
(2)若b
(Ⅰ)求数列an的通项公式;
(2)若b
(Ⅰ)因为{an}是等差数列,所以a3+a4=a2+a5=22又a3•a4=117
所以a3,a4是方程x2-22x+117=0的两根.又d>0,所以a3<a4.
所a3=9,a4=13,d=4,故a1=1,an=4n-3.
(Ⅱ)由(Ⅰ)可得Sn=
n(1+4n-3)
2=2n2-n,故bn=
2n2-n
n-
1
2=2n,
所以f(n)=
bn
(n+36)bn+1=
n
n2+37n+36=
1
n+
36
n+37≤
1
2
36+37=
1
49.
当且仅当n=
36
n,即n=6时,f(n)取得最大值
1
49.
所以a3,a4是方程x2-22x+117=0的两根.又d>0,所以a3<a4.
所a3=9,a4=13,d=4,故a1=1,an=4n-3.
(Ⅱ)由(Ⅰ)可得Sn=
n(1+4n-3)
2=2n2-n,故bn=
2n2-n
n-
1
2=2n,
所以f(n)=
bn
(n+36)bn+1=
n
n2+37n+36=
1
n+
36
n+37≤
1
2
36+37=
1
49.
当且仅当n=
36
n,即n=6时,f(n)取得最大值
1
49.
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