Discussion

Explanation:

u₁ = $\sqrt{3}$, 

u₂ = $\sqrt{3\sqrt{3}}$ = $\sqrt{3^1\times 3^{1/2}}$ = $3^{\frac{1}{2}}\times 3^{\frac{1}{4}}$ = $3^{\frac{1}{2}+\frac{1}{4}}$ = 3^{​​​​​​​3/4}, 

u₃ = $\sqrt{3\sqrt{3\sqrt{3}}}$ = $3^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}$ = 3^{​​​​​​​7/8}

∴ u_n = $3^{\frac{2^n-1}{2^n}}$

⇒ u₁₀ : u₉ = $3^{\frac{2^{10}-1}{2^{10}}}$ : $3^{\frac{2^9-1}{2^9}}$ 

= $3^{\frac{1023}{1024}-\frac{511}{512}}$ 

= $3^{\frac{1}{1024}}$ 

= $3^{\frac{1}{2^{10}}}$

Hence, option (d).

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