Question

$\sqrt{11\sqrt{11\sqrt{11...4\mathrm{terms}}}}=()?$

• A. $\sqrt[16]{{11}^{14}}$
• B. $\sqrt[16]{{11}^4}$
• C. $11$
• D. $\sqrt[16]{{11}^{15}}$

Answer 1

The correct option is D

$\sqrt[16]{{11}^{15}}$

Solution:

Solving the expression:

$$\begin{gathered} \sqrt{11\sqrt{11\sqrt{11...4\mathrm{terms}}}} \\ =\sqrt{11\sqrt{11\sqrt{11\sqrt{11}}}} \\ =\sqrt{11\sqrt{11\sqrt{11{\left(11\right)}^{\frac{1}{2}}}}}\mathbf{[}\mathbf{}\mathbf{\because }\sqrt{\mathbf{a}}\mathbf{=}{\mathbf{a}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{}\mathbf{]} \\ =\sqrt{11\sqrt{11(11\times {{\left(11\right)}^{\frac{1}{2}})}^{\frac{1}{2}}}} \\ =\sqrt{11{(11\times {11}^{\frac{1}{2}}{\times 11}^{\frac{1}{4}})}^{\frac{1}{2}}} \\ =(11\times {{11}^{\frac{1}{2}}\times {11}^{\frac{1}{4}}{\times 11}^{\frac{1}{8}})}^{\frac{1}{2}} \\ ={11}^{\frac{1}{2}}\times {11}^{\frac{1}{4}}{\times 11}^{\frac{1}{8}}\times {11}^{\frac{1}{16}}\left[\mathbf{\because }{\mathbf{}\mathbf{\left(}{\mathbf{a}}^{\mathbf{n}}\mathbf{\right)}}^{\mathbf{m}}\mathbf{=}\mathbf{}{\mathbf{a}}^{\mathbf{nm}}\right] \\ ={11}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{a}}^{\mathbf{m}}\mathbf{\times }\mathbf{}{\mathbf{a}}^{\mathbf{n}}\mathbf{=}\mathbf{}{\mathbf{a}}^{\mathbf{m}\mathbf{+}\mathbf{n}}\mathbf{]} \\ ={11}^{\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}[\mathrm{Taking}\mathrm{LCM}]} \\ ={11}^{\frac{15}{16}} \\ \mathbf{=}\sqrt[\mathbf{16}]{{\mathbf{11}}^{\mathbf{15}}}\mathbf{}\mathbf{}\mathbf{}\left[\mathbf{\because }\mathbf{}{\mathbf{a}}^{\frac{\mathbf{m}}{\mathbf{n}}}\mathbf{=}\mathbf{}\sqrt[\mathbf{n}]{{\mathbf{a}}^{\mathbf{m}\mathbf{}}}\right] \end{gathered}$$

Final answer: Hence, the correct option is (D).