1.44,24,13,7,4,2,( )
A.2 B.1 C.0 D.-1
2.2,8,24,64,( )
A.160 B.512 C.124 D.164
3.1,3,3,6,7,12,15,( )
A.17 B.27 C.30 D.24
4.45,29,21,17,15,( )
A.8 B.10 C.14 D.11
5.1,4,8,14, 24,42,( )
A.76 B.66 C.64 D.68
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1.B.【解析】仔細(xì)觀察,本題是連續(xù)的三項(xiàng)相減得到后一項(xiàng),所以括號(hào)中的數(shù)為1.
2.A.【解析】此題有相當(dāng)?shù)碾y度,初看似乎與冪有關(guān),或者呈直接的倍數(shù)關(guān)系,稍加假設(shè)驗(yàn)證,行不通。再看,項(xiàng)數(shù)不多,嘗試考察相連三數(shù)的關(guān)系,發(fā)現(xiàn)本數(shù)列其實(shí)是一個(gè)倍數(shù)關(guān)系的變形,(8-2)×4=24,(24-8)×4=64,所以下一個(gè)數(shù)是(64-24)×4=160.答案應(yīng)為A.
3.D.【解析】本題項(xiàng)數(shù)較多,分項(xiàng)錯(cuò)位考察,奇偶項(xiàng)單獨(dú)成數(shù)列,偶數(shù)項(xiàng)組成3,6,12,所以下一個(gè)為24.答案應(yīng)為D.
4.C.【解析】本題可依據(jù)常規(guī),把數(shù)列倒轉(zhuǎn),便于觀察,通過二級(jí)數(shù)列考察,相鄰兩數(shù)相減后形成一個(gè)比值為2的等比數(shù)列:2,4,8,16,所以答案應(yīng)選C.
5.A.【解析】根據(jù)前述一般規(guī)律,本題項(xiàng)數(shù)較多,采用兩次二級(jí)數(shù)列變形,相鄰兩數(shù)相減,得到一個(gè)公比為2的等比數(shù)列,答案應(yīng)選A.