Problem 1. Consider the numbers:

$$
x=1^{2}+3^{2}+5^{2}+\cdots+2015^{2} \text { and } y=2^{2}+4^{2}+6^{2}+\cdots+2016^{2}
$$

Using, if necessary, the relation $a^{2}-b^{2}=(a+b)(a-b)$, calculate the number: $A=(y-x)[y+x-(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+2015 \cdot 2016)]$
## Solution:

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=57&width=1692&top_left_y=819&top_left_x=153)

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=57&width=1698&top_left_y=871&top_left_x=156)

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=57&width=1698&top_left_y=919&top_left_x=150)

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=60&width=1692&top_left_y=963&top_left_x=153)

$y+x=1 \cdot(2-1)+2 \cdot(3-1)+3 \cdot(4-1)+\cdots+2016 \cdot(2017-1)$

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=49&width=1690&top_left_y=1066&top_left_x=154)

let: $p=1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+2015 \cdot 2016$

![](https://cdn.mathpix.com/cropped/2024_06_07_ba2c17359679d8926f12g-1.jpg?height=49&width=1692&top_left_y=1157&top_left_x=153)

$A=2017 \cdot 1008:(2016 \cdot 2017-2016 \cdot 1008)=1$..............................................................................................