Question

若(3-4x+x^2)(2+x-x^2)^3=a0+a1(1+x)+a2(1+x)^2+.+a8(1+x)^8,则a0+

Answer 1

对左、右两边进行分-------------------------------------------------------------------------------------左边=-x^8+7x^7-12x^6-14x^5+47x^4+3x^3-58x^2+4x+24-------------------------------------------------------------------------------------右边=a0+a1+a2+a3+a4+a5+a6+a7+a8+a1*x+2*a2*x+3*a3*x+4*a4*x+5*a5*x+6*a6*x+7*a7*x+8*a8*x+a2*x^2+3*a3*x^2+a3*x^3+6*a4*x^2+4*a4*x^3+a4*x^4+10*a5*x^2+10*a5*x^3+5*a5*x^4+a5*x^5+15*a6*x^2+20*a6*x^3+15*a6*x^4+6*a6*x^5+a6*x^6+21*a7*x^2+35*a7*x^3+35*a7*x^4+21*a7*x^5+7*a7*x^6+a7*x^7+28*a8*x^2+56*a8*x^3+70*a8*x^4+56*a8*x^5+28*a8*x^6+8*a8*x^7+a8*x^8-------------------------------------------------------------------------------------因此,若任意x等式均成立,则a0+a1+a2+a3+a4+a5+a6+a7+a8=24