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=______.
题型:安徽模拟难度:来源:
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=______. |
答案
∵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. |
举一反三
在(x-) 6的二项展开式中,常数项等于______. |
二项式(2-)6的展开式中含x项的系数是______. |
若(x+1)5-x5=a0+a1(x+4)4x+a2(x+1)3x2+a3(x+1)2x3+a4(x+1)x4,且a1(i=0,1,…,4)是常数,则a1+a3=______. |
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