Question

use Gauss's approach to find the follwoing sums (do not use formulas).
a... 1+2+3+4+...+999
b... 1+3+5+7+...+997
the sum of sequance a is...
the sum of sequance b is...

Answer 1

3+997 = 1000

Answer 2

i forgot how im suppose to solve this though

Answer 3

notice that you can sum up pairs of numbers such that their sum is 1000.
( 1 + 999 ) = 1000, ( 2 + 998 ) = 1000, ( 3 + 997 ) = 1000
count hot many time you can go this and multiply by 1000

Answer 4

but i dont know wat im supose to write for the answer im still lost lol

Answer 5

\[1\;\; \;\;\;+\;\; \;\;\;2\;\; \;\;\;+\;\; \;\;\;3\;\; \;\;\;+\;\; \;\;\;...........+996+997+998+999\]
\[999\;\; \;+\;\; \;998\;\; \;+\;\; \;997\;\; \;+\;\; \;\;\;.........+\;\; \;4\;\; \;+\;\; \;3\;\; \;+\;\; \;2\;\; \;+\;\; \;1\]
\[1000\;\;+\;\;1000\;\;+\;\;1000\;\;+\;\;.........+\;\;1000\;\;+\;\;1000\;\;+\;\;1000\;\;+\;\;1000\]
So, you've got 999 numbers, and all of them are 1000. So
\[\sum_{k=0}^{999} \;n= \frac{999*1000}{2}\]
Since you counted the sum twice, look at it carefully, I divided bu two.

Answer 6

499500??

Answer 7

yes