If x=1+\sqrt[6]{2}+\sqrt[6]{4}+\sqrt[6]{8}+\sqrt[6]{16}+\sqrt[6]{32}, then the value of (1+\frac{1}{x}) ^{24} is:

NIMCET 2024 Paper

Section: Mathematics

Question : 9 of 120

Marks: +1, -0

x=1+\;^{6}√{2}+\;^{6}√{4}+\;^{6}√{8}+\;^{6}√{16}+\;^{6}√{32}

, then the value of

(1+\;{1}/{x}) ^{24}

is:

Solution:

x=1+\;^{6}√{2}+\;^{6}√{4}+\;^{6}√{8}+\;^{6}√{16}+\;^{6}√{32}

Let us assume that

\;^{6}√{2}=a

, then

\;x=1+a+a^2+a^3+. . .+a^5=\;{a^6-1}/{a-1}

\;⇒ x=\;{2-1}/{\;^{6}√{2}-1}=\;{1}/{\;^{6}√{2}-1}

Hence

1+\;{1}/{x}=1+\;^{6}√{2}-1=\;^{6}√{2}

⇒ (1+\;{1}/{x}) ^{24}=(\;^{6}√{2})^{24}=16

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