5. Is the number

$$
\frac{1}{2+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\ldots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}
$$

rational? Justify!
The given sum can be written in the form

$$
A=\frac{1}{2 \cdot \sqrt{1}+1 \cdot \sqrt{2}}+\frac{1}{3 \cdot \sqrt{2}+2 \cdot \sqrt{3}}+\ldots+\frac{1}{100 \cdot \sqrt{99}+99 \cdot \sqrt{100}}
$$

The general term, starting from the second, is

$$
\begin{aligned}
\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} & =\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} \cdot \frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1) \sqrt{n}-n \sqrt{n+1}} \\
& =\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1)^{2} n-n^{2}(n+1)}=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n^{3}+2 n^{2}+n-n^{3}-n^{2}} \\
& =\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1} \\
& =\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
\end{aligned}
$$

(15 points)

Now we have

$$
\begin{aligned}
A & =1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{98}}-\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}} \\
& =1-\frac{1}{10}=\frac{9}{10}
\end{aligned}
$$

Thus, the given number is rational.

Note. The student can write out a few of the first terms in the sum and observe the given pattern.