Define a sequence langle a_n rangle by a_1 = | vedclass

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(B) Given $a_1 = 5$ and $a_n = a_1 a_2 \dots a_{n-1} + 4$ for $n > 1$.For $n > 2$,we have $a_n = (a_1 a_2 \dots a_{n-2}) a_{n-1} + 4$.Since $a_{n-1} =

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equals $1$

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(B) Given $a_1 = 5$ and $a_n = a_1 a_2 \dots a_{n-1} + 4$ for $n > 1$.For $n > 2$,we have $a_n = (a_1 a_2 \dots a_{n-2}) a_{n-1} + 4$.Since $a_{n-1} = a_1 a_2 \dots a_{n-2} + 4$,we have $a_1 a_2 \dots a_{n-2} = a_{n-1} - 4$.Substituting this into the expression for $a_n$:$a_n = (a_{n-1} - 4) a_{n-1} + 4 = a_{n-1}^2 - 4a_{n-1} + 4 = (a_{n-1} - 2)^2$.Thus,for $n > 2$,$\sqrt{a_n} = |a_{n-1} - 2| = a_{n-1} - 2$ (since $a_n$ is clearly increasing and $a_n > 2$).Therefore,$\lim_{n 	o \infty} \frac{\sqrt{a_n}}{a_{n-1}} = \lim_{n 	o \infty} \frac{a_{n-1} - 2}{a_{n-1}} = \lim_{n 	o \infty} \left( 1 - \frac{2}{a_{n-1}} ight)$.As $n 	o \infty$,$a_n 	o \infty$,so $\frac{2}{a_{n-1}} 	o 0$.The limit is $1 - 0 = 1$.

Define a sequence $\langle a_n \rangle$ by $a_1 = 5, a_n = a_1 a_2 \dots a_{n-1} + 4$ for $n > 1$. Then,$\lim_{n \to \infty} \frac{\sqrt{a_n}}{a_{n-1}}$

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