数列的问题

{an}是等差数列，{bn}是等比数列，且a2=b2 ＞0，a4=b4 ＞0，a1≠a4， b1＞0，试比较an与bn的大小，说明理由。

热心网友

{an}是等差数列，{bn}是等比数列，且a2=b2 ＞0，a4=b4 ＞0，a1≠a4， b1＞0，试比较an与bn的大小，说明理由。 解:①设等比数列公比为q,等差数列公差为d。∵b1＞0,b2＞0∴q＞0②∵a1≠a4,an不是常数列,∴a2≠a4即b2≠b4∴q≠1③设a2=b2=X＞0，a4=b4=Y＞0，X≠Y。a3=(X+Y)/2＞√XY=b3,∴a3＞b3又∵a1=2a2-a3=2X-(X+Y)/2=(3/2)X-(1/2)Yb1=(b2)^/b3=X^/(√XY)b1-a1=[X^/(√XY)+(1/2)Y]-(3/2)X=[X^/(2√XY)+X^/(2√XY)+(1/2)Y]-(3/2)X＞3{[X^/(2√XY)][X^/(2√XY)][(1/2)Y]}^(1/3)-(3/2)X=3{(X^3)/8}^(1/3)-(3/2)X=0b1-a1＞0a1＜b1当n＞4时,an＜bn下面运用归纳法证明这个结论④先证明b5＞a5,与b1＞a1相同a5=2a4-a3=2Y-(X+Y)/2=(3/2)Y-(1/2)Xb5=(b4)^/b3=Y^/(√XY)b5-a5=[Y^/(√XY)+(1/2)X]-(3/2)Y=[Y^/(2√XY)+Y^/(2√XY)+(1/2)X]-(3/2)Y＞3{[Y^/(2√XY)][y^/(2√XY)][(1/2)X]}^(1/3)-(3/2)Y=3{(y^3)/8}^(1/3)-(3/2)y=0∴b5＞a5⑤q＞1时,假设n=k(k＞4)时,即：ｂk＞ａk那么n=k+1时,由b5＞a5(且a4=b4=Y),Yq＞Y+d∴Y(q-1)＞dｂk+1-ｂk=qｂk-ｂk=(q-1)ｂk＞(q-1)ｂ4=(q-1)Y＞d=ａk+1-ａk∴ｂk+1-ｂk＞ａk+1-ａk∴ｂk+1-ａk+1＞ｂk-ａk＞0∴ｂk+1＞ａk+1综上所述:对于n＞4的自然数有an＜bn(q＞1)成立。⑥(0＜q＜1)的还用我证明吗?。