二项式定理一、复习目标:1.能利用二项式系数的性质求多项式系数的和与求一些组合数的和.2.能熟练地逆向运用二项式定理求和.3.能利用二项式定理求近似值,证明整除问题,证明不等式.二、课前预习:1.1003)32(的展开式中无理项的个数是(A)()A84()B85()C86()D872.设1510105)(2345xxxxxxf,则)(1xf等于(C)()A51x()B521x()C521x()D51x3.如果21872221221nnnnnCCC,则nnnnnCCCC210128.4.nnnnnCnCC11)1(3121121=11n.5.9)23(zyx展开式中含432zyx的项为43290720zyx.6.若1001002210100)1()1()1()21(xaxaxaax,则99531aaaa215100.四、例题分析:例1.已知}{na是等比数列,公比为q,设nnnnnnCaCaCaaS123121(其中Nnn,2),且nnnnnnCCCCS2101,如果1limnnnSS存在,求公比q的取值范围.解:由题意11nnqaa,nnS21,用心爱心专心)0()1()1(122111221111qqaCqCqqCaCqaCqaqCaaSnnnnnnnnnnnn∴nnnnnqaqaSS)21(2)1(111.如果1limnnnSS存在,则1|21|q或121q,∴212q或1q,故13q且0q.例2.(1)求多项式673410234)157()53()323(xxxxxx展开式各项系数和(2)多项式1000231000)22(xxxx展开式中x的偶次幂各项系数和与x奇次幂各项系数和各是多少?解:(1)设)()157()53()323()(2210673410234Nnxaxaxaaxxxxxxxfnn,其各项系数和为naaaa210.又∵102674102210316)157()53()3213()1(naaaaf,∴各项系数和为102316.(2)设30013001101000231000)22()(xaxaaxxxxxf,∴0)1(3001210aaaaf,2)1(3001210aaaaf,故1300131aaa,1300020aaa,∴)(xf展开式中x的偶次幂各项系数和为1,x奇次幂各项系数和为-1.例3.证明:(1)nknknkC032)(Nn;(2)12221223222120223222nnnnnnnnnCCCCCC)(Nn;用心爱心专心(3))(3)11(2Nnnn;(4)2222212)1(21nnnnnnnnCCC由(i)知用心爱心专心用心爱心专心