2019×1010=...

$$2019^2 - 2018^2 + 1017^2 - \cdots + 3^2 - 2^2 + 1^2 - 0^2 = 2019 + 2018 + 2017 + \cdots + 3 + 2 + 1 = \frac{2019(2020)}{2} = \boxed{2039190}$$

Noticing that $(n+1)^{2} - (n)^2 = n + n + 1$ We can rewrite $1^{2}+(-2^{2}+3^{2})+(-4^{2}+5^{2})...(-2018^{2}+2019^{2})$ as $=1+2+3+4+5.......+2018+2019$ $=2019*2020/2 = 2039190$ Edit: Snipped

We have: $1^2 - 2^2 + 3^2 - 4^2 + ... + 2017^2 - 2018^2 + 2019^2$ $=$ $- 3 - 7 - 11 - 15 - ... - 4035 + 2019^2$ $=$ $- \dfrac{4038 . 1009}{2} + 2019^2 = 2019^2 - 2019 . 1009 = 2019 . 1010 = 2039190$

$-x^2+(x+1)^2=-x^2+x^2+2x+1=2x+1=x+(x+1)$ $ \Rightarrow 1^2-2^2+3^2-...-2018^2+2019^2=1+2+3+...+2019$ $1+2+3+...+2019= \frac{2019 \cdot 2020}{2}=2019 \cdot 1010=2039190$ $\blacksquare$

My solution is the same as #3, so I'll keep this post for storage and not post anything.

Rijadinho wrote: $-x^2+(x+1)^2=-x^2+x^2+2x+1=2x+1=x+(x+1)$ $ \Rightarrow 1^2-2^2+3^2-...-2018^2+2019^2=1+2+3+...+2019$ $1+2+3+...+2019= \frac{2019 \cdot 2020}{2}=2019 \cdot 1010=2039190$ $\blacksquare$ I got the same solution

Similar to the Paranaense Math Olympiad 2024 P2 L2

Formula for the sum of alternating squares: $S=\frac{n(n+1)(2n+1)-8\lfloor \frac{n}{2} \rfloor (\lfloor \frac{n}{2} \rfloor +1)(2\lfloor \frac{n}{2} \rfloor+1)}{6}$ Therefore, $S=\frac{2019\cdot2020\cdot4039-8\cdot 1009\cdot 1010\cdot 2019}{6}=\boxed{2039190}$