Problem 9. Find the integer part of the expression

$a=\sqrt{1981+\sqrt{1981+\sqrt{1981+\ldots+\sqrt{1981+\sqrt{1981}}}},}$ if the number 1981 appears in it $n$ times ( $n \geqslant 2$ ).
Solution. Using a calculator, we obtain:

$$
\begin{aligned}
& a_{1}=\sqrt{1981} \approx 44.508426 \\
& a_{2}=\sqrt{1981+\sqrt{1981}} \approx 45.005648 \\
& a_{3}=\sqrt{1981+\sqrt{1981+\sqrt{1981}}} \approx 45.011173 \\
& a_{4}=\sqrt{1981+\sqrt{1981+\sqrt{1981+\sqrt{1981}}}} \approx \\
& \approx 45.011234 \\
& a_{5} \approx 45.011234
\end{aligned}
$$

From this, we hypothesize that the integer part of the number \( a \) is 45, i.e., \([a]=45\).

Let's verify this hypothesis as follows: Suppose,

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-027.jpg?height=114&width=1139&top_left_y=1716&top_left_x=364)

$$
\begin{aligned}
& \text{Then } \sqrt{1981+\ldots+\sqrt{1981+\sqrt{1981}}}>135 \text{ and } \\
& 1981+\ldots+\sqrt{1981+\sqrt{1981}}>135^{2}
\end{aligned}
$$

Continuing this process, we arrive at an obviously false inequality. Therefore, \([a]=45\).