高中数学：131(二项式定理)课件3(新人教B版选修2-3) 课件.ppt

,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,二项式定理(三)习题课,温故知新,2.,化简：,.,3.,展开式中含,x,3,项的系数为,_,。,的有理项,1.,求,：,1820,4.,的展开式中，第五项与第三项的二项式系,数之比为,14,：,3,，求展开式的常数项,温故知新,5.,展开式的各项系数和为,_,；,1,6.,展开式的二项式系数之和为,128,、那么展开式的项数是,；各项系数之和为,：,7.,的所有二项式的各项系数和是,；,2,n+1,-2,8.,则,-255,温故知新,1,、计算,0.997,3,的近似值（精确到,0.001,）,0.997,3,=(1-0.003),3,=130.003+30.003,2,0.003,3,130.003,=0.991,近似计算问题,练习,：求,2.998,6,的近似值（精确到小数点后第三位）；,2.998,6,=,(3-0.002),6,=3,6,63,5,0.002+153,4,0.002,2,203,3,0.002,3,+,3,6,63,5,0.002+153,4,0.002,2,=7292.916+0.00486,726.089,求：,11,2004,被,10,除的余数。,余数与整除问题,练：,55,10,被,8,除的余数,.,57,10,被,8,除的余数,.,求证：,55,55,+1,能被,8,整除；,因为,55,55,+1=,(561),55,+1=56M,1+1=,56M,所以,55,55,+1,能被,8,整除,.,余数与整除问题,3,、求证：,4,2n+1,+3,n+2,能被,13,整除；,4,2n+1,+3,n+2,=4,16,n,+9,3,n,=4,(13+3),n,+,93,n,=413M+43,n,+93,n,=413M+133,n,所以,4,2n+1,+3,n+2,能被,13,整除,.,题组四,(,求值、等式与不等式证明问题,),求证：,