2. The numbers are given:

$$
\begin{aligned}
a & =1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\ldots+\frac{1}{1+2+3+\ldots+2015} \\
b & =\frac{1}{2}+\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2014}\right)+\left(\frac{2}{2}+\frac{2}{3}+\frac{2}{4}+\ldots+\frac{2}{2014}\right) \\
& +\left(\frac{3}{3}+\frac{3}{4}+\ldots+\frac{3}{2014}\right)+\ldots+\left(\frac{2013}{2013}+\frac{2013}{2014}\right)+\left(\frac{2014}{2014}\right)
\end{aligned}
$$

Determine which of the numbers $\sqrt{a-1}, \sqrt{b}, \sqrt{2 a b}$ is rational, justifying your answer.
Solution.

$$
\begin{aligned}
a & =\frac{2}{1 \cdot 2}+\frac{2}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\ldots+\frac{2}{2015 \cdot 2016} \\
a & =2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\ldots+\frac{1}{2015}-\frac{1}{2016}\right) \\
a & =2\left(\frac{1}{1}-\frac{1}{2016}\right)=\frac{2015}{1008}
\end{aligned}
$$

and

$$
\begin{aligned}
& b=\frac{1}{2}+1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots+\frac{1+2+\ldots+2014}{2014} \\
& b=\frac{1}{2}+\frac{2}{2}+\frac{2 \cdot 3}{2 \cdot 2}+\frac{3 \cdot 4}{2 \cdot 3}+\ldots+\frac{2014 \cdot 2015}{2 \cdot 2014} \\
& b=\frac{1+2+3 \ldots+2015}{2}=\frac{2015 \cdot 2016}{2 \cdot 2}=2015 \cdot 504
\end{aligned}
$$

Thus, $\sqrt{a-1}=\sqrt{\frac{1007}{1008}}$. Since $\frac{1007}{1008}$ is an irreducible fraction, and the last digits of the numerator and denominator are 7 and 8, respectively, it follows that neither the denominator nor the numerator are perfect squares, so $\left.\sqrt{a-1}=\frac{\sqrt{1007}}{\sqrt{1008}} \notin \mathbb{Q} . \mathbf{( 1 p}\right)$

$\sqrt{b}=\sqrt{\frac{2015 \cdot 2016}{4}}=\frac{\sqrt{2015 \cdot 2016}}{2}$. Since 2015.2016 is the product of consecutive numbers, it is not a perfect square, so $\sqrt{b} \notin \mathbb{Q}$. (1p)

$\sqrt{2 a b}=\sqrt{2 \cdot \frac{2015}{1008} \cdot 2015 \cdot 504}=\sqrt{2015^{2}}=2015 \in \mathbb{Q}$. (1p)