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[學(xué)科競賽]競賽輔導(dǎo)-數(shù)列一等差數(shù)列與等比數(shù)列-文庫吧資料

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【正文】 )(( bcbcabab ??????abbcbcab??????))((又 ))()(())(())((11cbacabbcbccbacabcbac ?????????????18 ))(()(acabbc????acab ????11?cb ?1 , ac ?1 , ba ?1 成等差數(shù)列; a, b, c ? 互不相等 cb ?1 , ac ?1 , ba ?1?不成等比數(shù)列; 選 C 不是常數(shù)列; cb ?1 , ac ?1 , ba ?119 例 4 Ten players took part in roundrobin touynament ( every player must play against every other player exactly once). There were no draws in this tournament. Suppose that the first player won games,the second player won games , the third player won games and so on .The value of is= 1x2x 3x1021 xxx ??? ?45 20 ? ?na 1a1 2nnS nSn? ?? na例 已知等差數(shù)列 的首項(xiàng)為 ,前 n項(xiàng)的和 成立,則 = . 為 Sn使等式 解:1 2nnS nSn? ??由 nnSSannn 21 ???? ?nSann 2111 ???? ? 12 ??? nn naS11 )( ??? nn naaan11 aaa nn ??? ?? ?na是等差數(shù)列 1ad ??111 )1( naanaa n ?????1na21 *11( 1 ) ( ) , 0 ,nnn a n a n N a?? ? ? ?{}na若數(shù)列 則 (A)是等比數(shù)列但不是等差數(shù)列 (B) 是等差數(shù)列但不是等比數(shù)列 (C)是等差數(shù)列也是等比數(shù)列 (D)不是等差數(shù)列也不是等比數(shù)列 ( ) 22 4 23 }{ na 1221 ????? nnaaa ??1a???? 22221 naaa ?例 數(shù)列 中, 則 _____________, ____________. 1 312 2 ?n24 11 1ba?? 33bS{}nab數(shù)列{ an}是各項(xiàng)均為正數(shù)的等差數(shù)列,前 n項(xiàng)的 和為 Sn。 結(jié)合 函數(shù) ( 或不等式 等 ) 的思想、觀點(diǎn)和方法分析解決有關(guān)數(shù)列的問題 。 2022 2022 2022:/8320王新敞源頭學(xué)子小屋 (差 ) m+n=p+q, m,n,p,q∈ N* daa nn ??? 1q不可以 是 0 d可以 是 0 等比中項(xiàng) abG ???等差中項(xiàng) ba
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