Compute the value

The sum works out to 1001.

sumfor(n, 1, 1010,  (2*n - 1) - (2*n))== - 1010 + 2021==1011

OR:

2021 + ∑ [ (2n - 1) - (2n), n, 1, 1010] ==1011

Compute the value of 

1-2+3-4+...+2019-2020+2021

(1+3+5+ ..... +2021) - ( 2+4+6+ ..... +2020)

these are just arithmetic progresstions

The sum of an AP = $\frac{n}{2}(a+L)$

There are 2021 terms in total  half that 1010.5   so there are 1010 even numbers and 1011 odd numbers

(1+3+5+ ..... +2021) - ( 2+4+6+ ..... +2020)

$=[\frac{1011}{2}(1+2021)]-[\frac{1010}{2}(\color{red}{2} \color{black}+2020)]\\ =[1011*(1011)]-[505*(2022)]\\ =1022121 - 1021110\\ =1011$

I found my careless error, the corrected number is in red.

I am sorry to say this and I apologize but 1011 and 1516 aren't the answers and idk how and what it is.

Nvm scratch that the answer is 1011.

Sorry!!