20.已知數(shù)列的首項(xiàng).前n項(xiàng)和. 查看更多

 

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(本題滿分14分)已知數(shù)列的首項(xiàng),….

(1)證明:數(shù)列是等比數(shù)列;

(2)數(shù)列的前項(xiàng)和

 

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(本題滿分14分)已知數(shù)列的首項(xiàng),通項(xiàng)為常數(shù)),且成等差數(shù)列.

(1)求的值;

(2)數(shù)列的前項(xiàng)的和.

 

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(本題滿分14分)已知數(shù)列是首項(xiàng)為1公差為正的等差數(shù)列,數(shù)列是首項(xiàng)為1的等比數(shù)列,設(shè),且數(shù)列的前三項(xiàng)依次為1,4,12,

(1)求數(shù)列、的通項(xiàng)公式;

(2)若等差數(shù)列的前n項(xiàng)和為Sn,求數(shù)列的前項(xiàng)的和Tn

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(本題滿分14分)已知數(shù)列是首項(xiàng)為1公差為正的等差數(shù)列,數(shù)列是首項(xiàng)為1的等比數(shù)列,設(shè),且數(shù)列的前三項(xiàng)依次為1,4,12,

(1)求數(shù)列、的通項(xiàng)公式;

(2)若等差數(shù)列的前n項(xiàng)和為Sn,求數(shù)列的前項(xiàng)的和Tn

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(本題滿分14分)已知數(shù)列是首項(xiàng)為1公差為正的等差數(shù)列,數(shù)列是首項(xiàng)為1的等比數(shù)列,設(shè),且數(shù)列的前三項(xiàng)依次為1,4,12,

(1)求數(shù)列、的通項(xiàng)公式;

(2)若等差數(shù)列的前n項(xiàng)和為Sn,求數(shù)列的前項(xiàng)的和Tn

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1-10.CDBBA   CACBD

11. 12. ①③④   13.-2或1  14.   15.2  16.  17..

18.

解:(1)由已知            7分

(2)由                                                                   10分

由余弦定理得                          14分

 

19.(1)證明:∵PA⊥底面ABCD,BC平面AC,∴PA⊥BC,                                  3分

∵∠ACB=90°,∴BC⊥AC,又PA∩AC=A,∴BC⊥平面PAC.                             5分

(2)解:過(guò)C作CE⊥AB于E,連接PE,

∵PA⊥底面ABCD,∴CE⊥面PAB,

∴直線PC與平面PAB所成的角為,                                                    10分

∵AD=CD=1,∠ADC=60°,∴AC=1,PC=2,

中求得CE=,∴.                                                  14分

 

20.解:(1)由①,得②,

②-①得:.                              4分

(2)由求得.          7分

   11分

.                                                                 14分

 

21.解:

(1)由得c=1                                                                                     1分

,                                                         4分

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    市一次模文數(shù)參答―1(共2頁(yè))

                                                                                            5分

    (2),時(shí)取得極值.由.                                                                                          8分

    ,,∴當(dāng)時(shí),

    上遞減.                                                                                       12分

    ∴函數(shù)的零點(diǎn)有且僅有1個(gè)     15分

     

    22.解:(1) 設(shè),由已知,

    ,                                        2分

    設(shè)直線PB與圓M切于點(diǎn)A,

                                                     6分

    (2) 點(diǎn) B(0,t),點(diǎn),                                                                  7分

    進(jìn)一步可得兩條切線方程為:

    ,                                   9分

    ,

    ,,                                          13分

    ,又時(shí),,

    面積的最小值為                                                                            15分

     

     

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