#Same sequence 1

$a_1 = -2520 \\ a_2 = -4540 = -2520 \color{#69047E} {- 2020} \\$ $a_3 = -6559 = -2520 - 2020 - 2019 = -2520 \color{#69047E} {- (2) 2020 }\color{#3D99F6} { + 1 } \\$ $a_4 = -8577 = -2520 - 2020 - 2019 - 2018 = -2520 -\color{#69047E} { (3) 2020} \color{#3D99F6} { + 1 + 2 }\\$ $a_5 = -8577 = -2520 - 2020 - 2019 - 2018 -2017 = -2520 - \color{#69047E} {(4) 2020} \color{#3D99F6} {+ 1 + 2 + 3}$

$a_n = -2520 - \color{#69047E} {2020 (n-1)} \color{#3D99F6} {+ \cfrac{(n-2)(n-1)}{2} }$

For $n = 4050$

$\begin{aligned} a_{4050} &= -2520 - 2020 (4050-1) + \cfrac{(4050-2)(4050-1)}{2} \\ &= -2520 -2020(4050-1) + 2024 (4050-1) \\ &= -2520 + 4 (4049) \\ &= \boxed{13676} \end{aligned}$

$-25\underbrace{20,-45}_{2020}\underbrace{40,-65}_{2019}\underbrace{59,-85}_{2018}\underbrace{77,-105}_{2017}94,...$

We note that the difference between consecutive terms starts from $2020$ and decreases by $1$ consecutively. Let the $n$ th term be $a_n$ and $a_0=-499$ . Then we have:

$\begin{aligned} \\ a_1 & = - 499 - 2021 = -2520 \\ a_2 & = - 499 - 2021 - 2020 = - 4520 \\ a_3 & = - 499 - 2021-2020-2019 = - 6559 \\ \cdots & = \cdots \\ \implies a_n & = - 499 - \underbrace{(2021 + 2020 + 2019 + \cdots + (2022-n))}_{n \text{ terms}} \\ & = - 499 - \frac {n(4043-n)}2 \\ \implies a_{4050} & = -499 - \frac {4050(-7)}2 = \boxed{13676} \end{aligned}$

f(x) = -2020x-500 for x=1,2

Now, f(x) = k(x-1)(x-2)-2020x-500 for any x=n

Here, f(3) = 2k-2020×3-500

Or, -6559 = 2k-6560

So, k=0.5

Then, f(n) = 0.5(n-1)(n-2)-2020n-500

So, f(4050) = 13676