高考數(shù)學(xué)沖刺小題狂練07數(shù)列

2nn222nn22數(shù)

列(建議用時(shí)：40分鐘)一、選擇題．S是差數(shù)列{}前n項(xiàng)和，若S＝，則a等()．n74A8C．

B．7D．．已知等比數(shù)列{}前項(xiàng)依次為－1，＋1，＋，則a＝)．nA

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B．4·

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1C．4·

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D．4·

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1．已知數(shù)列{a}前項(xiàng)和S＝－＋2則數(shù){}通項(xiàng)公式為(．nnnAa＝2n－3nC．＝≥

B．a(chǎn)＝2n＋3nD．＝≥2．等差數(shù)列{a}前項(xiàng)和為S，且＋a＝，＝，則＝()nnA8C．

B．9D．11．在等比數(shù){}，a，是程－x＋2＝兩根，則a的值是()．n86A±C.2

B－D．．已知等比數(shù)列{}公為正數(shù)，且a＝，＝，則a＝()．n3952

C.2

D．列{}由數(shù)組成的等比數(shù)為前項(xiàng)和a＝1＝＝()n4C.

．設(shè)等差數(shù){}前和為S，a＞0＋＞，a＜，則滿(mǎn)足＞的最大n13106n自然數(shù)值為)．A6C．

B．7D．13/4

2*4*2*4*已知各項(xiàng)不為0的等差數(shù)列{a}足－a＋a＝0{}等數(shù)列＝，n478n77則b等于()．2811A1C．

B．2D．10已知函數(shù)f)＝－)＋10(為常數(shù))，若數(shù)列{}足＝(n)(nN)且＝，nn則數(shù)列{}100項(xiàng)和為()．nA39400C．

B－D．78二、填空題．等差數(shù)列{}，＋＝，a＋＝，則＋＝n1256910S12設(shè)等比數(shù){}公比q＝，前項(xiàng)和為，的為_(kāi)_______nna313知正數(shù)b的比中項(xiàng)是且＝＋＝＋則m的小值是．a(chǎn)b14已知數(shù){}足＝n

＋＋3＋?，則數(shù)列{}前n項(xiàng)和為_(kāi)．a(chǎn)nn115整數(shù)數(shù){}足＝－(∈N，若此數(shù)列前800項(xiàng)和是2013前n1n項(xiàng)的和是，則其前014項(xiàng)和．/4

1742n1222622533222274351742n122262253322227435參考答案一、選擇題7．解析答案D4．解析(a1)(aaaa64·12

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n

答案B．解析1≥3.n11nn1nC.

答案．解析

aa(1)35

9.8

答案B．解析

00

aa2.答4868

C解析{}aaan952

2a6

qa2.5a1q5

答案．解析q(0)1a1.S7234[131760，aS3qq214答案B．解析a00aaaa1676310aa2aS012.答案113712nC解析d2303d2a)047877a2ab{}){}qba77n7bbb)答案D281110析∵f3m)103f(n2?1101×10100×10×100400.二、填空題/4

答案B

41423nnn41423nnn．解析aaa4a4()5156210656()8aaa8d26.答案612191061512解析aaq.4q4

答案

113解析b2abmb，nb12()()≥ab5a2“”m答案3?114解析ann2aa114n

111＋－?3n1

1

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答案

n＋2．解析

aaaaaaaa?31432536461aaa?}aa2831014aa