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

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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

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

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