题目内容
(2012•安徽模拟)设x6=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+a4(x-1)4+a5(x-1)5+a6(x-1)6,则a3=
20
20
.分析:由于x=(x-1)+1,x6=[(x-1)+1]6=
+
•(x-1)1+
•(x-1)2+…+
•(x-1)6,与已知对比可得a3=
,从而可求得a3.
| C | 0 6 |
| C | 1 6 |
| C | 2 6 |
| C | 6 6 |
| C | 3 6 |
解答:解:∵x6=[(x-1)+1]6=
+
•(x-1)1+
•(x-1)2+…+
•(x-1)6
=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+a4(x-1)4+a5(x-1)5+a6(x-1)6,
∴a3=
=20.
故答案为:20.
| C | 0 6 |
| C | 1 6 |
| C | 2 6 |
| C | 6 6 |
=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+a4(x-1)4+a5(x-1)5+a6(x-1)6,
∴a3=
| C | 3 6 |
故答案为:20.
点评:本题考查二项式定理的应用,观察分析得到an=
是关键,考查分析与转化的能力,属于中档题.
| C | n 6 |
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