数列an中,前n项和sn=an^2+bn,其中a,b是常数,且a＞0,a+b＞1,n∈N*

s(n)=n^2*a+bn,
a(1)=s(1)=a+b,
a(n+1)=s(n+1)-s(n)=(2n+1)*a+b=2(n+1)a-a+b,
a(n)=2na-a+b=a+b+(n-1)*(2a).
a(n)=a+b+(n-1)*(2a)>=a+b>1
a(n+1)=a+b+n*(2a)=a+b+(n-1)*(2a)+2a>a+b+(n-1)*(2a)=a(n)>1
c(n)=log[a(n)a(n+1)]=log[a(n)]+log[a(n+1)],
c(n+1)=log[a(n+1)]+log[a(n+2)]
c(n+1)-c(n)=log[a(n+2)]-log[a(n)]=log[a(n+2)]-log[a(n+1)]+log[a(n+1)]-log[a(n)]
=log[a(n+2)/a(n+1)]+log[a(n+1)/a(n)]
>log[1] + log[1]
=0
c(n+1)>c(n).
a(1)=1, a(n)>0.
0=(n+1)[a(n+1)]^2 - n[a(n)]^2 + a(n+1)a(n)
这个, 不会...