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∴...