20xx年高考數(shù)學(xué)總復(fù)習(xí)12命題及其關(guān)系充分條件與必要條件限時(shí)練習(xí)新人教版-資料下載頁(yè)
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【正文】 ≥0,bn+3bn+2≥0,∴由⑤得bn+1bn＝0(n＝1,2,3,…).由此不妨設(shè)bn＝d3(n＝1,2,3,…),則ana n+2＝d3(常數(shù)).由此＝an+2an+1+3aa+2＝4an+2an+13d3,從而+1＝4an+1+2an+23d3＝4a n+1+2an5d3.兩式相減,得+1＝2(an+1an)2d3,因此an+1an＝(+1)+d3＝d2+d3(常數(shù))(n＝1,2,3…),∴數(shù)列{}是等差數(shù)列.證法二:令A(yù)n＝a n+1an,由bn≤b n+1,知ana n+2≤a n+1a n+3,從而a n+1an≥a n+3a n+2,即An≥A n+2(n＝1,2,3…).由＝an+2a n+1+3a n+2,c n+1＝a n+1+2a n+2+3a n+3,得c n+1＝(a n+1an)+2(a n+2a n+1)+3(a n+3a n+2),即An+2A n+1+3A n+2＝d2時(shí) ⑥由此得A n+2+2A n+3+3An+4＝d2. ⑦⑥⑦,得(AnA n+2)+2(A n+1A n+3)+3(A n+2An+4)＝0. ⑧∵AnA n+2≥0,A n+1A n+3≥0,A n+2A n+4≥0,∴由⑧,得AnA n+2＝0(n＝1,2,3,…).于是由⑥,得4An+2A n+1＝An+2A n+1+3A n+2＝d2, ⑨從而2An+4A n+1＝4A n+1+2A n+2＝d2. ⑩由⑨和⑩,得4An+2A n+1＝2An+4A n+1,故A n+1＝An,即a n+2a n+1＝a n+1an(n＝1,2,3,…),∴數(shù)列{}是等差數(shù)列
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