高中数学若(x+1)^5-x^5=a0+a1(x+1)^4x+a2(x+1)^3x^2+a3(x+1)^2x^3+a4(x+1)x^4且ai均为常数,则a1+a3=

2024-12-15 17:00:31
推荐回答(1个)
回答1:

据题意:(x+1)5-x5=a0+a1(x+4)4x+a2(x+1)3x2+a3(x+1)2x3+a4(x+1)x4中,
左式=(x+1)5-x5=C50x5+C51x4+…+C55x0-x5=C51x4+C52x3+C53x2+C54x+1,
分析可得左式中常数项为1,右式中常数项为a0,则a0=1;
左式中x的1次项为5,右式中x的1次项为C51,C51=a1即a1=5
左式中x的2次项为C52,右式中x的2次项为C41a1+a2,则C52=C41a1+a2即4a1+a2=10
解可得,a2=-10
左式中x的3次项为C53,右式中x的3次项为C42a1+C31a2+a3
则C53=C42a1+C31a2+a3即10=6a1+3a2+a3

解可得a3=10
所以a1+a3=15
故答案为15.