解答
展开 (2a3+a3)6
解答
64a18+576a29+2160a11+4320a15+2916a+4860a4+a3729
求解步骤
(2a3+a3)6
使用二项式定理: (a+b)n=i=0∑n(in)a(n−i)bia=2a3,b=a3
=i=0∑6(i6)(2a3)(6−i)(a3)i
展开求和
=0!(6−0)!6!(2a3)6(a3)0+1!(6−1)!6!(2a3)5(a3)1+2!(6−2)!6!(2a3)4(a3)2+3!(6−3)!6!(2a3)3(a3)3+4!(6−4)!6!(2a3)2(a3)4+5!(6−5)!6!(2a3)1(a3)5+6!(6−6)!6!(2a3)0(a3)6
化简 0!(6−0)!6!(2a3)6(a3)0:64a18
化简 1!(6−1)!6!(2a3)5(a3)1:576a229
化简 2!(6−2)!6!(2a3)4(a3)2:2160a11
化简 3!(6−3)!6!(2a3)3(a3)3:(a)34320a9
化简 4!(6−4)!6!(2a3)2(a3)4:4860a4
化简 5!(6−5)!6!(2a3)1(a3)5:(a)52916a3
化简 6!(6−6)!6!(2a3)0(a3)6:a3729
=64a18+576a229+2160a11+(a)34320a9+4860a4+(a)52916a3+a3729
(a)34320a9=a4320a8
(a)52916a3=a2916a
=64a18+576a229+2160a11+a4320a8+4860a4+a2916a+a3729
合并 64a18+576a229+2160a11+a4320a8+4860a4+a2916a+a3729:64a18+576a229+2160a11+108a(40a7+27)+4860a4+a3729
=64a18+576a229+2160a11+108a(40a7+27)+4860a4+a3729
展开 108a(40a7+27):4320a215+2916a
=64a18+576a229+2160a11+4320a215+2916a+4860a4+a3729
化简 64a18+576a229+2160a11+4320a215+2916a+4860a4+a3729:64a18+576a29+2160a11+4320a15+2916a+4860a4+a3729
=64a18+576a29+2160a11+4320a15+2916a+4860a4+a3729