Question

1.多项式f(x)=Cn,1(x-1)+Cn,2(x-1)^2+Cn,3(x-1)^3+.+Cn,n(x-1)^n的展开

Answer 1

1.6（x-1）^6

=6[(x-1)^2]^3

=6(x^2+1-2x)^3

=6(x^2+1-2x)[x^4+(2-4x)x+1+4x^2-4x]

=6(x^2+1-2x)(x^4+2x-4x^2+1+4x^2-4x)

=6(x^2+1-2x)(x^4-2x+1)

=6(x^6-2x^3+x^2+x^4-2x+1-2x^5+4x^2-2x)

=6(x^6-2x^5+x^4-2x^3+5x^2-4x+1)

=6x^6-12x^5+6x^4-12x^3+30x^2-24x+6

12(x-1)^12

=12[(x-1)^6]^2

=12[x^12+(2x^4-4x^5-4x^3+10x^2-8x+2)x^6+4x^10+x^8+4x^6+25x^4+16x^2+1-8x-5(2-8x)x^2-(10x^2-8x+2)x^3+(10x^2-4x^3-8x+2)x^4-2(2x^4-4x^3+10x^2-8x+2)x^5]

2.原式=（1-x^2)^5(1+x^2+2x)

=(1+x^4-2x^2)^2(1-x^2)(1+x^2+2x)

=(1+2x^4-4x^2+x^8+4x^4-4x^6)(1-x^2)(1+x^2+2x)

=(x^8-4x^6+6x^4-4x^2+1)(1-x^2)(1+x^2+2x)

=(x^8-4x^6+6x^4-4x^2+1-x^10+4x^8-6x^6+4x^4-x^2)(1+x^2+2x)

=(-x^10+5x^8-10x^6+10x^4-5x^2+1)(1+x^2+2x)

=-x^10+5x^8-10x^6+10x^4-5x^2+1-x^12+5x^10-10x^8+10x^6-5x^4+x^2-2x^11+10x^9-20x^7+20x^5-10x^3+2x

=-x^10+5x^8+10x^4-5x^2+1-x^12+5x^10-10x^8-5x^4+x^2-2x^11+10x^9-20x^7+20x^5-10x^3+2x

含x^6项的系数是0.