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a(n)=aq^(n-1),2a(3)=2aq^2=a(1)+a(2)=a+aq,0=a[2q^2-q-1]=a(2q+1)(q-1), a=0,或, q=-1/2,或q=1. 因公比不为0,所以, a不为0.(1)若q=1,则a(n)=a, s(n)=na.b(n)=2+(n-1)q=2+n-1=n+1,s(n)-b(n)=na-n-1=n(a-1)-1,n>=2时,(1.1)若a>3/2,则a-1>1/2, n(a-1)>1, s(n)-b(n)=n(a-1)-1>0, s(n)>b(n).(1.2)若a<=1,则s(n)-b(n)=n(a-1)-1<0, s(n)<b(n).(1.3)若1<a<=3/2, ... |
