高中數(shù)學(xué)數(shù)列知識(shí)點(diǎn)解析

1、數(shù)數(shù)列列知識(shí)要點(diǎn)知識(shí)要點(diǎn)知識(shí)要點(diǎn)知識(shí)要點(diǎn)1.⑴等差、等比數(shù)列：等差數(shù)列等比數(shù)列等差數(shù)列等比數(shù)列定義daann???1)0(1???qqaann遞推公式daann???1；mdaanmn???qaann1??；mnmnqaa??通項(xiàng)公式dnaan)1(1???11??nnqaa（01?qa）中項(xiàng)2knknaaA????（0??knNkn?）)0(?knknknknaaaaG??????（0??knNkn?）前n項(xiàng)和)(21nnaanS??

2、dnnnaSn2)1(1??????????????????)2(111)1(111qqqaaqqaqnaSnnn重要性質(zhì))(qpnmNqpnmaaaaqpnm???????)(qpnmNqpnmaaaaqpnm???????數(shù)列數(shù)列的定義數(shù)列的有關(guān)概念數(shù)列的通項(xiàng)數(shù)列與函數(shù)的關(guān)系項(xiàng)項(xiàng)數(shù)通項(xiàng)等差數(shù)列等差數(shù)列的定義等差數(shù)列的通項(xiàng)等差數(shù)列的性質(zhì)等差數(shù)列的前n項(xiàng)和等比數(shù)列等比數(shù)列的定義等比數(shù)列的通項(xiàng)等比數(shù)列的性質(zhì)等比數(shù)列的前n項(xiàng)和②112???

3、?nnnaaa(2?n，011???nnnaaa)①注①：i.acb?，是a、b、c成等比的雙非條件，即acb?a、b、c等比數(shù)列.ii.acb?（ac＞0）→為a、b、c等比數(shù)列的充分不必要.iii.acb??→為a、b、c等比數(shù)列的必要不充分.iv.acb??且0?ac→為a、b、c等比數(shù)列的充要.注意：任意兩數(shù)a、c不一定有等比中項(xiàng)，除非有ac＞0，則等比中項(xiàng)一定有兩個(gè).③nncqa?(qc為非零常數(shù)).④正數(shù)列na成等比的充要條

4、件是數(shù)列nxalog（1?x）成等比數(shù)列.⑷數(shù)列na的前n項(xiàng)和nS與通項(xiàng)na的關(guān)系：?????????)2()1(111nssnasannn[注]：①????danddnaan??????111（d可為零也可不為零→為等差數(shù)列充要條件（即常數(shù)列也是等差數(shù)列）→若d不為0，則是等差數(shù)列充分條件）.②等差na前n項(xiàng)和ndandBnAnSn?????????????????22122→2d可以為零也可不為零→為等差的充要條件→若d為零，則是等

5、差數(shù)列的充分條件；若d不為零，則是等差數(shù)列的充分條件.③非零常數(shù)列既可為等比數(shù)列，也可為等差數(shù)列.（不是非零，即不可能有等比數(shù)列）2.①等差數(shù)列依次每k項(xiàng)的和仍成等差數(shù)列，其公差為原公差的k2倍...232kkkkkSSSSS??；②若等差數(shù)列的項(xiàng)數(shù)為2????Nnn，則，奇偶ndSS??1??nnaaSS偶奇；③若等差數(shù)列的項(xiàng)數(shù)為?????Nnn12，則??nnanS1212???，且naSS??偶奇，1??nnSS偶奇得到所求項(xiàng)數(shù)到