题目内容
(2012•桂林模拟)对任意实数x,有x3=a0+a1(x-2)+a2(x-2)2+a3(x-2)3,则a2=( )
分析:根据题意,将x3变形为[(x-2)+2]3,由二项式定理可得x3=[(x-2)+2]3=C30(x-2)023+C3122(x-2)+C3221(x-2)2+C3320(x-2)3,又由题意,可得a2=C3221,计算可得答案.
解答:解:根据题意,x3=a0+a1(x-2)+a2(x-2)2+a3(x-2)3,
而x3=[(x-2)+2]3=C30(x-2)023+C3122(x-2)+C3221(x-2)2+C3320(x-2)3,
则a2=C3221=6;
故选B.
而x3=[(x-2)+2]3=C30(x-2)023+C3122(x-2)+C3221(x-2)2+C3320(x-2)3,
则a2=C3221=6;
故选B.
点评:本题考查二项式定理的应用,关键是将x3变形为[(x-2)+2]3,进而由二项式定理将其展开.

练习册系列答案
相关题目