+0

# Compute the value

+1
188
5
+27

Compute the value of

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

May 12, 2022
edited by IcanhelpandIneedhelp  May 12, 2022

#1
+1

The sum works out to 1001.

May 12, 2022
#2
+1

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

OR:

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

May 12, 2022
#3
+118459
+1

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.

May 13, 2022
edited by Melody  May 19, 2022
#4
+27
0

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

May 17, 2022
#5
+27
+1

Nvm scratch that the answer is 1011.

Sorry!!

May 17, 2022