大侠们,等差数列中,公差d>0,前n项和为Sn,且满足a2*a4=45,a1+a5=14,求（1）数列{an}通项公式及Sn.（2）令bn=1/(an^2-1),数列{Cn}满足c1=-1/4,Cn+1-Cn=bn,求数列{Cn}通项公式Cn.(3)求f(n)=n/9-bn/cn的最小值.

大侠们,等差数列中,公差d>0,前n项和为Sn,且满足a2*a4=45,a1+a5=14,求（1）数列{an}通项公式及Sn.（2）令bn=1/(an^2-1),数列{Cn}满足c1=-1/4,Cn+1-Cn=bn,求数列{Cn}通项公式Cn.(3)求f(n)=n/9-bn/cn的最小值.
大侠们,
等差数列中,公差d>0,前n项和为Sn,且满足a2*a4=45,a1+a5=14,求（1）数列{an}通项公式及Sn.
（2）令bn=1/(an^2-1),数列{Cn}满足c1=-1/4,Cn+1-Cn=bn,求数列{Cn}通项公式Cn.
(3)求f(n)=n/9-bn/cn的最小值.

大侠们,等差数列中,公差d>0,前n项和为Sn,且满足a2*a4=45,a1+a5=14,求（1）数列{an}通项公式及Sn.（2）令bn=1/(an^2-1),数列{Cn}满足c1=-1/4,Cn+1-Cn=bn,求数列{Cn}通项公式Cn.(3)求f(n)=n/9-bn/cn的最小值.
（1）、a2*a4=(a1+d)(a1+3d)=45,
a1+a5=a1+a1+4d=14,
——》d=2,a1=3,
——》an=a1+(n-1)d=2n+1,
Sn=na1+n(n-1)d/2=n^2+2n；
（2）、bn=1/(an^2-1)=1/4n(n+1)=1/4[1/n-1/(n+1)],
Cn+1-Cn=bn=1/4[1/n-1/(n+1)],
——》Cn-C(n-1)=1/4[1/(n-1)-1/n],
.
C2-C1=1/4(1-1/2),
——》C(n+1)-C1=1/4[1-1/(n+1)],
——》C(n+1)=1/4-1/4(n+1)+C1=-1/4(n+1),
——》Cn=-1/4n；
（3）、f(n)=n/9-bn/cn=n/9-[1/4n(n+1)]/(-1/4n)=n/9+1/(n+1),
——》f(n)=(n+1)/9+1/(n+1)-1/9>=2v[(n+1)/9*1/(n+1)]-1/9=2/3-1/9=5/9,
当且仅当(n+1)/9=1/(n+1)时等号成立,即n=2时,f(n)的最小值f(2)=5/9.

(1)
an =a1+(n-1)d
a2.a4=45
(a1+d)(a1+3d) =45 (1)
a1+a5=14
2a1+4d = 14
a1 = 7-2d (2)
sub (2) into (1)
(7...

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(1)
an =a1+(n-1)d
a2.a4=45
(a1+d)(a1+3d) =45 (1)
a1+a5=14
2a1+4d = 14
a1 = 7-2d (2)
sub (2) into (1)
(7-d)(7+d)=45
d^2 = 4
d =2
a1 = 3
an = 3+(n-1)2 = 2n +1
Sn = (2n+1+3)n/2
=(n+2)n
(2)
bn =1/((an)^2-1)
= 1/( (2n+1)^2 -1)
= 1/[ (2n+2)(2n) ]
= (1/4)[1/n - 1/(n+1)]

c(n+1) -cn = bn
= (1/4)[1/n - 1/(n+1)]
c(n+1) + (1/4)[1/(n+1)] = cn +(1/4)(1/n)
{c(n+1) + (1/4)[1/(n+1)]}- [cn +(1/4)(1/n)]=0
cn +(1/4)(1/n) = c1+ 1/4
=0
cn = -1/(4n)
(3)
f(n)=n/9-bn/cn
= n/9 - (1/[4n(n+1)])/[-1/(4n)]
= n/9 + 1/(n+1)
consider
f(x) = (1/9)[x+ 9/(x+1)]
f'(x) = (1/9)[ 1 - 9/(x+1)^2 ] =0
(x+1)^2 -9 =0
x^2+2x-8 =0
(x+4)(x-2) =0
x=-4 or 2 (min)
min f(n) =f(2)
= 2/9 + 1/3
= (2+3)/9
=5/9

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