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Mathematics 22 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

so like (1+999) +(2 +998) +(3+997)... and 1+(3+997)+(5+995)+(7+993)... ?

OpenStudy (anonymous):

what do i do next?

OpenStudy (anonymous):

1+999 =?

OpenStudy (anonymous):

you need to realize they are all in groups of 1000 so you count up the number of groups you can make, then add that to whatever number you cant make a group with

OpenStudy (anonymous):

i know that they are in a group of 1000..but do i add up 997? im confused?

OpenStudy (anonymous):

just focus on the first problem you can make groups of 1000 up to 501 and 499 999-501 groups of 1000 999-501=498 this leaves 500 so 498 groups of 1000 = 498*1000=498000 add the 500 to it you get 498500

OpenStudy (anonymous):

for the second problem, you need to realize the only number you can't make a group with is 1 then just count up all the groups of 1000 you can make or you could make groups of 998 but personally 1000 is easier to calculate

OpenStudy (anonymous):

249001? is that for the second problem

OpenStudy (anonymous):

no idea, didnt calculate it

OpenStudy (anonymous):

because you dont have 1 and 999 as a pair i think there are 248 pairs instead of 249 but im just guessing

OpenStudy (anonymous):

thanks

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