# PROBLEM 1

a) Show that: $A=\sqrt{1+3+5+\ldots+2015} \in Q$

b) If the real numbers $a$ and $b$ satisfy the relation: $a^{2}+b^{2}-4 \sqrt{3} a-6 \sqrt{2} b+30=0$, then calculate:

$$
\mathrm{E}=\left(2 \mathrm{a}^{-1}+3 \mathrm{~b}^{-1}\right)\left(\frac{1}{b^{-1}}-\frac{1}{a^{-1}}\right)
$$
## Solution:

a) $1+3+5+\ldots+2015=(1+2+3+4+\ldots+2015)-(2+4+6+\ldots+2014)=$ $\qquad$
$=2015 \cdot 1008-1007 \cdot 1008=1008^{2}$

$A=\sqrt{1+3+5+\ldots+2015}=1008 \in Q \ldots \ldots . . . . . . .1$ point
b) $(a-2 \sqrt{3})^{2}+(b-3 \sqrt{2})^{2}=0 \quad$................ 1 point
$a-2 \sqrt{3}=0 \Rightarrow a=2 \sqrt{3}, \quad b-3 \sqrt{2}=0 \Rightarrow b=3 \sqrt{2}$ $\qquad$ 1 point

$E=\left(2 \cdot \frac{1}{2 \sqrt{3}}+3 \cdot \frac{1}{3 \sqrt{2}}\right)(3 \sqrt{2}-2 \sqrt{3})$ $\qquad$ 1 point

$E=1$

1 point