Question

设实数a不等于0且函数f(x)=a(x^2+1)-(2x+1/a)有最小值—1

Answer 1

f(x)=a(x^2+1)-(2x+1/a),a=1

f(x)=x^2+1-2x+1=x^2-2x

Sn=f(n)=n^2-2n

S(n-1)=(n-1)^2-2(n-1)=n^2-4n+3

an=Sn-S(n-1)=n^2-2n-(n^2-4n+3)=2n-3

数列｛an}是以-1为首项,公差为2的等差数列

a1=-1,a2=1

那么a2,a4,a6.a2n就是以1为首项,公差为4的等差数列

bn=(a2+a4+…a2n)/n=[n+n(n-1)*4/2]/n=2n-1

所以数列｛bn}是等差数列,其首项为1,公差为2.

证毕.