[3X—1/(X)^(2/3)]展开式中各项系数和为128,则展开式中1/X^3的系数是???????
来源:百度知道 编辑:UC知道 时间:2024/06/11 08:35:56
[3X—1/(X)^(2/3)]展开式中各项系数和为128,则展开式中1/X^3的系数是???????
详细过程!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
是的!!→[3X—(1/X^(2/3))]^n
详细过程!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
是的!!→[3X—(1/X^(2/3))]^n
题目是→[3X—(1/X^(2/3))]^n←吗?
将X=1代入,则可得各项和为2^n=128.∴n=7.得原式为[3X—(1/X^(2/3))]^7
由二项式定理展开式得通项为:7Cr×(3X)^(7-r)×[-X^(-2/3)]^r.
整理得3^(7-r)×(-1)^r×7Cr×X^[7-(5/3)r].①∴令7-(5/3)r=-3.
得r=6,代入①得3^(7-6)×(-1)^6×7C6×X^(-1/3)=21/X^3.
∴1/X^3的系数为21
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