The sequence is defined recursively. Write the first four terms.
-$$a _ { 1 } = \sqrt { 6 } ; a _ { n } = \sqrt { 6 a _ { n - 1 } }$$
A) $\sqrt { 6 } , 6,6 \sqrt { 6 } , 36$

B) $\sqrt { 6 } , \sqrt { \sqrt { 6 } } , \sqrt { \sqrt { \sqrt { 6 } } } , \sqrt { \sqrt { \sqrt { \sqrt { 6 } } } }$

C) $\sqrt { 6 } , 6 \sqrt { 6 } , 36 \sqrt { 6 } , 216 \sqrt { 6 }$

D) $\sqrt { 6 } , \sqrt { 6 \sqrt { 6 } } , \sqrt { 6 \sqrt { 6 \sqrt { 6 } } } , \sqrt { 6 \sqrt { 6 \sqrt { 6 \sqrt { 6 } } } }$

Question

The sequence is defined recursively. Write the first four terms.
-$$a _ { 1 } = \sqrt { 6 } ; a _ { n } = \sqrt { 6 a _ { n - 1 } }$$
A) $\sqrt { 6 } , 6,6 \sqrt { 6 } , 36$

B) $\sqrt { 6 } , \sqrt { \sqrt { 6 } } , \sqrt { \sqrt { \sqrt { 6 } } } , \sqrt { \sqrt { \sqrt { \sqrt { 6 } } } }$

C) $\sqrt { 6 } , 6 \sqrt { 6 } , 36 \sqrt { 6 } , 216 \sqrt { 6 }$

D) $\sqrt { 6 } , \sqrt { 6 \sqrt { 6 } } , \sqrt { 6 \sqrt { 6 \sqrt { 6 } } } , \sqrt { 6 \sqrt { 6 \sqrt { 6 \sqrt { 6 } } } }$

Quizplus · Accepted Answer

The sequence starts with $a_1 = \sqrt{6}$. Each subsequent term is the square root of 6 times the previous term, leading to a pattern of nested square roots.

The Sequence Is Defined Recursively a1=6;an=6an−1a _ { 1 } = \sqrt { 6 } ; a _ { n } = \sqrt { 6 a _ { n - 1 } }a1​=6​;an​=6an−1​​

The Sequence Is Defined Recursively $a _ { 1 } = \sqrt { 6 } ; a _ { n } = \sqrt { 6 a _ { n - 1 } }$