已知等比數列{an}的首項a1>0,公比為q>-1,且q≠0。數列{bn}的通項bn=an+1(n+1是下標)+an+2(n+2是下標),它們前n項的和分別記作An和Bn,試比較An和Bn的大小。

熱心網友

Bn=A(n+1)+A(n+2)-a1-a1-a2=2An+2a(n+1)+a(n+2)-2a1-a2=2An+a1(2q^n+q^(n+1)-2-q)==2An+a1(q^n-1)(2+q)=1)q≥1,An≥0,a1(q^n-1)(2+q)≥0,Bn≥2AnAn.2)1q-1Bn=2An+a1(q^n-1)(2+q)=a1(q^n-1)[2/(q-1)+(2+q)]==a1[(q^n-1)/(q-1)][q(q+1)/(q+2)]=An[q(q+1)/(q+2)]An=a1[(q^n-1)/(q-1)]0ⅰ)1q0,q(q+1)/(q+2)An.1q-1,和-1