1.
bn=a1+a2+a3...an\n
nbn=a1+a2+a3...an=n^3
an=n^3-(n-1)^3=3n^2-3n+1
2.
令a1+a2+a3...an=Sn
bn=b+(n-1)d
bn=a1+a2+a3...an\n=Sn/n
Sn=nb+n(n-1)d
a1=S1=b
an=Sn-S(n-1)=b+(n-1)*2d
an是b為首項2d為公差的等差數(shù)列
x1x4=m,x2x3=n
x1+x4=2,x2+x3=2
x1=1/4,x4=7/4
x2=3/4,x3=5/4
m=x1x4=7/16,n=x2x3=15/16
由于mn地位均等故或者n=x1x4=7/16,m=x2x3=15/16
不論怎樣|m-n|=1/2
an=a1+(n-1)d
a11=a1+10d>=0
a12=a1+11d