sumfor(n, 1, 1010, (2*n - 1) - (2*n))== - 1010 + 2021==1011
OR:
2021 + ∑ [ (2n - 1) - (2n), n, 1, 1010] ==1011
+118608 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.
+27 I am sorry to say this and I apologize but 1011 and 1516 aren't the answers and idk how and what it is.
