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If $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\ldots n}}}$ times, then which one of the following is true?
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1323 Upvotes
Verified Answer
The correct answer is:
$a_n < 4, \forall n \geq 1$
On limiting case $a_n < a_{\infty}$
Now, $\quad a_{\infty}=\sqrt{7+a_{\infty}}$
$$
\Rightarrow a_{\infty}^2-a_{\infty}-7=0 \Rightarrow a_{\infty}=\frac{1+\sqrt{29}}{2} \quad\left[\because a_{\infty}>0\right]
$$
So, $\quad a_n < \frac{1+\sqrt{29}}{2}, \forall n \geq 1$
$$
\therefore \quad a_n < 4, \forall n \geq 1 .
$$
Now, $\quad a_{\infty}=\sqrt{7+a_{\infty}}$
$$
\Rightarrow a_{\infty}^2-a_{\infty}-7=0 \Rightarrow a_{\infty}=\frac{1+\sqrt{29}}{2} \quad\left[\because a_{\infty}>0\right]
$$
So, $\quad a_n < \frac{1+\sqrt{29}}{2}, \forall n \geq 1$
$$
\therefore \quad a_n < 4, \forall n \geq 1 .
$$
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