题目内容
若(x+a)(x+1)(x+2)(x+3)展开式中含x3项的系数是17,则a的值( )
| A.10 | B.11 | C.12 | D.13 |
(x+a)(x+1)(x+2)(x+3)
=[x2+(a+1)x+a](x2+5x+6)
=x4+(a+1)x3+ax2+5x3+5(a+1)x2+5ax+6x2+6(a+1)x+6a
=x4+(a+6)x3+(6a+11)x2+(11a+6)x+6a.
∴a+6=17,
解得a=11.
故选B.
=[x2+(a+1)x+a](x2+5x+6)
=x4+(a+1)x3+ax2+5x3+5(a+1)x2+5ax+6x2+6(a+1)x+6a
=x4+(a+6)x3+(6a+11)x2+(11a+6)x+6a.
∴a+6=17,
解得a=11.
故选B.
练习册系列答案
相关题目