数列题帮忙

显示全部楼层 · 2009-3-28 20:25:30

{an}为公比不为1的正项等比数列，则[ ]
A.a1+a8>a4+a5
B.a1+a8=a4+a5
D.a1+a80,q>1a1+a8-(a4+a5) =a1+a1q^7-(a1*q^3+a1*q^4) =(a1-a1q^3)+(a1q^7-a1q^4) =a1(1-q^3)+a1q^4(q^3-1) =a1(1-q^3)-a1q^4(1-q^3) =a1(1-q^3)(1-q^4) =a1[(1-q)(1+q+q^2)][(1+q^2)(1+q)(1-q)] =a1(1-q)^2(1+q)(1+q^2)[(1/2+q)^2+3/4] 由于a1>0,q>1则:a1(1-q)^2(1+q)(1+q^2)[(1/2+q)^2+3/4]>0则:a1+a8>a4+a5...

千问 · 2009-3-28 20:25:30

因为公比不为1的正项等比数列，所以设公比为q>1a1+a8=a1+a1q^7=a1(1+q^7)a4+a5=a1q^3+a1q^4=a1q^3(1+q)a1(1+q^7)/a1q^3(1+q)=(1+q^7)/[q^3(1+q)]>q^7/[q^3(1+q)]=q^4/(1+q)>1 (因为q>1)所以，a1(1+q^7)>a1q^3(1...

千问 · 2009-3-28 20:25:30

a1+a8－（a4+a5）＝a1+a1q^7-a1q^3-a1q^4=a1(1+q^7-q^3-q^4)=a1[(q^7-q^3)+(1-q^4)]=a1[q^3(q^4-1)-(q^4-1)]=a1[(q^4-1)(q^3-1)]当q>1时，q^4-1>0,q^3-1>0,a1>0,a1+a8－（a4+a5）>0当qa4+a5...