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The last term in the binomial expansion of ${(\sqrt{2} - \frac{1}{\sqrt{2}})}^{n}$ is ${(\frac{1}{(3) 9^{\frac{1}{3}}})}^{\log 3}$ then the 5th term from the beginning is

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10C6

2 10C4

12 10C4

−10C6

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detailed solution

Correct option is B

nCn−12n=3−1−2/3log3⁡8=3−5log3⁡2=125⇒ n=10 ∴ 5th term from the beginning =10C4(2)10−4−124=2 10C4

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The last term in the binomial expansion of2−12n is 1(3)913log⁡3then the 5th term from the beginning is

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The last term in the binomial expansion of ${(\sqrt{2} - \frac{1}{\sqrt{2}})}^{n}$ is ${(\frac{1}{(3) 9^{\frac{1}{3}}})}^{\log 3}$ then the 5th term from the beginning is