1. Given the numbers $a=\sqrt{1+3+5+7+9+11}-\sqrt{1+3+5+7+9}, b=\sqrt{1+3+5+\ldots+4029}$ and $c=\sqrt{1+3+5+\ldots+4027}$. Calculate:
a) $a^{2015}$
b) $(b-c)^{2015}$
c) $(b+c)^{a-1}$
## Solution:

a) $1+3+5+7+9+11=1+2+3+4+\ldots+11-(2+4+\ldots+10)=\frac{11 \cdot 12}{2}-2 \cdot \frac{5 \cdot 6}{2}=6^{2}, \ldots \ldots \ldots \ldots .1 \text{p}$ $1+3+5+7+9=1+2+3+4+\ldots+9-(2+4+\ldots+8)=5^{2}, a=\sqrt{6^{2}}-\sqrt{5^{2}}=6-5=1 \ldots \ldots \ldots \ldots . .1 \text{p}$

![](https://cdn.mathpix.com/cropped/2024_06_07_954bb18b226b938f15feg-2.jpg?height=60&width=1665&top_left_y=1255&top_left_x=184)
b) using a) we get $1+3+5+\ldots+4029=1+2+3+4+\ldots+4029-(2+4+\ldots+4028)=2015^{2}$, so $b=\sqrt{2015^{2}}=2015$ $.1 \text{p}$ analogously $c=\sqrt{2014^{2}}=2014$ $1 \text{p}$

$(b-c)^{2015}=1^{2015}=1$ $1 \text{p}$

c) $(b+c)^{a-1}=4029^{0}=1$ $1 \text{p}$