(x- 3x^2 + x^26)^10 的展开式中,x^13的系数是`?

(x- 3x^2 + x^26)^10 的展开式中,x^13的系数是`?请写出过程`谢谢
2025-01-24 05:26:49
推荐回答(3个)
回答1:

看成10个(x-3x^2+x^26)相乘,每个式子里含有一个x一次方,一个x二次方,一个x二十六次方,10个式子凑出x的十三次方:7个x,3个x^2
想这一项为:C(10)3 * x^7 * (-3x^2)^3 = -3240x^13
系数为-3240

回答2:

观察各指数可知 指数皆>O 目标指数13<26
所以式孑中指数为26那项可不看 即(x-3x~2)~10
由二项式定理知含x项为 (C10 3)*x~7*(-3x~2)~3
系数为10*9*8/(1*2*3)*(-3)~3=-3240

回答3:

x^26一项不需要考虑
(x-3x^2)^10=x^10(1-3x)^10
x^13对应的系数为C(10,3)*(-3)^3*1^7=-3240

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