Exercise 11. For all integers $n \geqslant 0$, we define

$$
U_{n}=\frac{2 n+1-2 \sqrt{n(n+1)}}{\sqrt{n+1}-\sqrt{n}}
$$

Calculate $\left(\mathrm{U}_{0}+\cdots+\mathrm{U}_{41}\right)^{2}$.
## Solution to Exercise 11

We start by calculating the first terms of the sequence $\left(\mathrm{U}_{n}\right)$.

For $\mathrm{n}=0$, we have $\mathrm{U}_{0}=1$, for $\mathrm{n}=1, \mathrm{U}_{1}=\frac{3-2 \sqrt{2}}{\sqrt{2}-1}=\frac{(3-2 \sqrt{2})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}-1$.

For $\mathrm{n}=2, \mathrm{U}_{2}=\frac{5-2 \sqrt{6}}{\sqrt{3}-\sqrt{2}}$, by multiplying the numerator and denominator by $\sqrt{3}+\sqrt{2}$, as $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1$, we get $\mathrm{U}_{2}=(5-2 \sqrt{6})(\sqrt{3}+\sqrt{2})=5 \sqrt{3}-4 \sqrt{3}-5 \sqrt{2}-6 \sqrt{2}=\sqrt{3}-\sqrt{2}$. From these two initial values, we can reasonably conjecture that $U_{n}=\sqrt{n+1}-\sqrt{n}$ for all integers $n$.

Let's verify this. The fact that $U_{n}=\sqrt{n+1}-\sqrt{n}$ is equivalent to the equality $2 n+1-2 \sqrt{n(n+1)}=(\sqrt{n+1}-\sqrt{n})^{2}$, which corresponds to a notable identity. We have thus obtained a simple formula for $\mathrm{U}_{\mathrm{n}}$.

Now let's finish the exercise. Given the expression for $\mathrm{U}_{\mathrm{n}}$, the sum $\mathrm{U}_{0}+\mathrm{U}_{1}+\ldots+\mathrm{U}_{41}$ is a telescoping sum. In particular,

$$
\mathrm{U}_{0}+\cdots+\mathrm{U}_{41}=\sqrt{1}-\sqrt{0}+\cdots+\sqrt{42}-\sqrt{41}=\sqrt{42}
$$

so $\left(\mathrm{U}_{0}+\cdots+\mathrm{U}_{41}\right)^{2}=42$

Grader's Comment: The problem is very well solved.