1、第5讲计数原理与二项式定理1.(2019江苏,22,10分)设(1+x)n=a0+a1x+a2x2+anxn,n4,nN*,已知a32=2a2a4.(1)求n的值;(2)设(1+3)n=a+b3,其中a,bN*,求a2-3b2的值.2.(2019苏中、苏北七大市三模,22)设Pn=i=02n(-1)iC2ni,Qn=j=12n(-1)jjC2nj.(1)求2P2-Q2的值;(2)化简nPn-Qn.答案精解精析1.解析(1)因为(1+x)n=Cn0+Cn1x+Cn2x2+Cnnxn,n4,所以a2=Cn2=n(n-1)2,a3=Cn3=n(n-1)(n-2)6,a4=Cn4=n(n-1)(n-2
2、)(n-3)24.因为a32=2a2a4,所以n(n-1)(n-2)62=2n(n-1)2n(n-1)(n-2)(n-3)24.解得n=5.(2)由(1)知,n=5.(1+3)n=(1+3)5=C50+C513+C52(3)2+C53(3)3+C54(3)4+C55(3)5=a+b3.解法一:因为a,bN*,所以a=C50+3C52+9C54=76,b=C51+3C53+9C55=44,从而a2-3b2=762-3442=-32.解法二:(1-3)5=C50+C51(-3)+C52(-3)2+C53(-3)3+C54(-3)4+C55(-3)5=C50-C513+C52(3)2-C53(3)3
3、+C54(3)4-C55(3)5.因为a,bN*,所以(1-3)5=a-b3.因此a2-3b2=(a+b3)(a-b3)=(1+3)5(1-3)5=(-2)5=-32.2.解析(1)由于P2=1C40-1C41+1C42-1C43+1C44=53,Q2=-1C41+2C42-3C43+4C44=103,所以2P2-Q2=0.(2)设T=nPn-Qn,则T=nC2n0-nC2n1+nC2n2-+nC2n2n-1C2n1+2C2n2-3C2n3+2nC2n2n=nC2n0-n-1C2n1+n-2C2n2-n-3C2n3+-nC2n2n.因为C2nk=C2n2n-k,所以T=nC2n2n-n-1C2n2n-1+n-2C2n2n-2-n-3C2n2n-3+-nC2n0=-nC2n0-1-nC2n1+2-nC2n2-3-nC2n3+nC2n2n.+得,2T=0,即T=nPn-Qn=0,所以nPn-Qn=0.
