Question

设函数a≠0,且函数f(x)=a(x^+1)-(2x+1/a)有最小值-1,设数列an的前n项和sn=f(n),令bn=

Answer 1

f(x)=a(x^2+1)-(2x+1/a),a>0最小值为a-2/a=-1,a=1

f(x)=x^2-2x,sn=f(n)=n^2-2n

n>=2,an=sn-s(n-1)=2n-3

n=1,a1=s1=-1

an=2n-3,a2n=4n-3,a2+a4+……+a2n=(2n-1)n

bn=(a2+a4+……+a2n)/n=2n-1

b(n+1)-bn=2

所以bn是等差数列