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題目是→[3X—(1/X^(2/3))]^n←嗎?將X=1代入,則可得各項和為2^n=128.∴n=7.得原式為[3X—(1/X^(2/3))]^7由二項式定理展開式得通項為:7Cr×(3X)^(7-r)×[-X^(-2/3)]^r.整理得3^(7-r)×(-1)^r×7Cr×X^[7-(5/3)r].①∴…
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