1-(x-1)+(x-1)^2-(x-1)^3+.+(x-1)^20的展开式中x^3项的系数过程谢谢.~答案为-5985
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![1-(x-1)+(x-1)^2-(x-1)^3+.+(x-1)^20的展开式中x^3项的系数过程谢谢.~答案为-5985](/uploads/image/z/5343293-29-3.jpg?t=1-%28x-1%29%2B%28x-1%29%5E2-%28x-1%29%5E3%2B.%2B%28x-1%29%5E20%E7%9A%84%E5%B1%95%E5%BC%80%E5%BC%8F%E4%B8%ADx%5E3%E9%A1%B9%E7%9A%84%E7%B3%BB%E6%95%B0%E8%BF%87%E7%A8%8B%E8%B0%A2%E8%B0%A2.%7E%E7%AD%94%E6%A1%88%E4%B8%BA-5985)
1-(x-1)+(x-1)^2-(x-1)^3+.+(x-1)^20的展开式中x^3项的系数过程谢谢.~答案为-5985
1-(x-1)+(x-1)^2-(x-1)^3+.+(x-1)^20的展开式中x^3项的系数
过程谢谢.~
答案为-5985
1-(x-1)+(x-1)^2-(x-1)^3+.+(x-1)^20的展开式中x^3项的系数过程谢谢.~答案为-5985
看成以1为首项,(1-x)为公比的等比数列!
用求和公式!
得[1-(1-x)^21]/x
其中x^3项的系数为
4
C 21 =[(-1)^17]*(21*20*19*18)/(1*2*3*4)
=-5985
(-1)^n(x-1)^n展开式中x^3项的系数为
-C(n 3)=-[n(n-1)(n-2)/6]
=-(1/6)(n³-3n²+2n)
系数和=-(1/6)[(1³+2³+...+20³)-3(1²+2²+...+20²)+2(1+2+...+20)]
=-(1/6)[20²×21²/4-3×20×21×41/6+20×21]
=-5985
嗯,答案是正确的,是-5985。
1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20
=(1+(x-1)^21)/(1+(x-1))=(1+(x-1)^21)/x=1/x+(x-1)^21/x
((x^4)*((-1)^17)(21*20*19*18)/(1*2*3*4))/x=-5985x^3
1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20的各项恰好构成了首项为1、公比为-(x-1)的等比数列,故
1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20=1*{1-[-(x-1)]^21}/[1+(x-1)]=[1-(1-x)^21]/x
要求1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1...
全部展开
1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20的各项恰好构成了首项为1、公比为-(x-1)的等比数列,故
1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20=1*{1-[-(x-1)]^21}/[1+(x-1)]=[1-(1-x)^21]/x
要求1-(x-1)+(x-1)^2-(x-1)^3+.......+(x-1)^20的展开式中x^3项的系数就是求[1-(1-x)^21]/x展开式中x^4项的系数
又[1-(1-x)^21]/x展开式中x^4项的系数为-C4 21=-(21*20*19*18)/(4*3*2*1)=-5985
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