Question

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

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an通项公式
（2）通过bn=Sn/n+c，构造一个新的数列｛bn｝，是否存在一个非实零数c，使｛bn｝也为等差数列
（3）求f[n]=bn/（n+25）b（n+1）的最大值

Answer 1

1、由a2×a3=45,a1+a4=14得：
{（a1+d)(a1+2d)=45 a1+a1+3d=14
解之得a1=13 d=-4(舍去） 或a1=1 d=4
故{an}是以1为首项，公差为4的等差数列
an=1+(n-1)*4=4n-3
2、Sn=(1+4n-3)*n/2=n(2n-1)
故：bn=Sn/(n+c)=[n(2n-1)]/(n+c)
b(n+1)=S(n+1)/(n+1+c)=[(n+1)(2n+1)]/(n+1+c)
b(n+1)-bn=[(n+1)(2n+1)]/(n+1+c)- [n(2n-1)]/(n+c)
=[(n+c)(n+1)(2n+1)-(n+1+c)n(2n-1)]/[(n+c)(n+1+c)]
=[2n^2+2(1+2c)n+c]/[n^2+(1+2c)n+c^2+c]
=2*[n^2+(1+2c)n+c/2]/[n^2+(1+2c)n+c^2+c]
若要{bn}为等差数列，b(n+1)-bn应为常数，即公差为常数。
则必须c/2=c^2+c c(c+1/2)=0
由于c不为零，故c=-1/2
些时公差为d=b(n+1)-bn=2 首项为b1=S1/(1+c)=1/(1/2)=2,，
通项为：bn=Sn/(n+c)=[n(2n-1)]/(n-1/2)=[2n(2n-1)]/(2n-1)=2n
数列{bn}为等差数列。
3、f(n)=bn/[(n+25)*b(n+1)]=2n/[(n+25)*2*(n+1)]=n/[(n+1)(n+25)]
f(n+1)=b(n+1)/[(n+25)*b(n+2)]=2(n+1)/[(n+1+25)*2*(n+2)]= (n+1)/[(n+2)(n+26)]
f(n+1)-f(n)=(n+1)/[(n+2)(n+26)]-n/[(n+1)(n+25)]
=[(n+1)(n+1)(n+25)-n(n+2)(n+26)]/[(n+1)(n+2)(n+25)(n+26)]
=(-n^2-n+25)/[(n+1)(n+2)(n+25)(n+26)]
由于(n+1),(n+2),(n+25),(n+26)全为正数，故f(n+1)-f(n)的符号由二次函数-n^2-n+25确定。
令g(x)=-x^2-x+25 其对称轴为x=-1/2，故当x取正整数时单调递减且g(1)=-1-1+25=23>0
当g(x)=0时 x=(-1-√101)/2 或者x=(-1+√101)/2
而4<(-1+√101)/2 <5
因此x取正整数， 1≤x≤4时 g(x)>0； 当x≥5时 g(x)<0。
综上分析 当n≤4时 f(n+1)-f(n)>0 ；当n≥5时f(n+1)-f(n)<0
故f(5)是最大值，此时：
f(5)=b5/[(n+25)*b6]=(2*5)/[(5+25)*2*6]=1/36

Answer 2

(1)
an=a1+(n-1)d

a2.a3=4
(a1+d)(a1+2d) =45 (1)

a1+a4=14
2a1+3d=14 (2)
sub (2) into (1)
[(14-3d)/2 + d].[(14-3d)/2+2d ]= 45
(14-d).(14+d)= 180
196-d^2=180
d= 4
from (2), a1= 1
an =1+4(n-1) = 4n-3
(2)
Sn = (2n-1)n
bn = Sn/(n+ c)
c= -1/2
bn = 2n
=> {bn} 为等差数列

(2)
f(n)=bn/[(n+25).[b(n+1) ]

= n/[(n+25)(n+1)]
let
f(x) = x/[(x+25)(x+1)]
f'(x) =[(x+25)(x+1) -x(2x+26) ] /[(x+25)(x+1)]^2
= (-x^2+25)/[(x+25)(x+1)]^2
f'(x) =0
x=5
max f(n) = f(5)
=5/[(5+25)(5+1)]
= 1/36