If $x = 1 + 2^{1/6} + (2^{1/6})^2 + (2^{1/6})^3 + (2^{1/6})^4 + (2^{1/6})^5$, then the value of $(1 + 1/x)^{24}$ is

✅ Correct Answer: 16

Explanation

Given $x = 1 + 2^{1/6} + (2^{1/6})^2 + (2^{1/6})^3 + (2^{1/6})^4 + (2^{1/6})^5$ This is a geometric progression with first term $1$ and common ratio $2^{1/6}$ $x = (1 - (2^{1/6})^6) / (1 - 2^{1/6})$ Since $(2^{1/6})^6 = 2$ $x = 1 / (2^{1/6} - 1)$ Now $1 + 1/x = 2^{1/6}$ Therefore $(1 + 1/x)^{24} = (2^{1/6})^{24} = 2^4 = 16$

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If $x = 1 + 2^{1/6} + (2^{1/6})^2 + (2^{1/6})^3 + (2^{1/6})^4 + (2^{1/6})^5$, then the value of $(1 + 1/x)^{24}$ is | ACME Academy