-(x＋2)²＋16(x-1）² a(n-1)²＋4a(n-1)＋4a
问题描述：
-(x＋2)²＋16(x-1）² a(n-1)²＋4a(n-1)＋4a
答
-(x+2)²+16(x-1)²=16(x-1)²-(x+2)²=[4(x-1)]²-(x+2)² 运用平方差公式=[4(x-1)+(x+2)] [4(x-1)-(x+2)]=(4x-4+x+2)(4x-4-x-2)=(5x-2)(3x-6)=3(5x-2)(x-2)a(n-1)²+4a(n-1)+4a=a[(n-...