Question

NEED HELP!!

What is 1+2+3+4+5+6.......999

Answer 1

do you just need the sum?

Answer 2

do you mean 1+2+3.... all the way to 998+999??

Answer 3

1+2+3+____+999
it is an arithmetic progression series

Answer 4

common difference = a_2 - a_1 = 2-1 = 1

Answer 5

go to wolframalpha.com and type in 1+2+3+4+5+6+...+999=

Answer 6

so here i go with the formula .. :
\[\large{s=\frac{n}{2}[2a+(n-1)d]}\]

Answer 7

given a = 1 , n =999
\[\large{\frac{999}{2}[2+998]}\]
\[\large{\frac{999}{2}[1000]}\]= 499500

Answer 8

got it @vmathhelp ? ..

Answer 9

Yes, and thank you

Answer 10

Here is a nice way to figure it out:
Say you add up the numbers 1 through 100
so you have
1+2+3+.....+98+99+100
So you add up the first and the last 1+100=101
add up 2+99=101
add up 3+98=101 you get the point, you have 50 pairs of these right?
so you have 50(101)=5,050
and i found this online:
http://mathforum.org/library/drmath/view/57919.html

Answer 11

isn't it 500,499

Answer 12

no problem ..

Answer 13

no @vmathhelp it will be 499,500

Answer 14

999 * 500 = 499500

Answer 15

I didn't post answer i posted the logic behind it

Answer 16

http://answers.yahoo.com/question/index?qid=20080818114809AAJFb6n

Answer 17

500,500

Answer 18

notice: +100....

Answer 19

your series dont contain 100 and so on

Answer 20

formulas aren't helpful if you don't understand their derivation (where they're coming from) i hardly memorized any formulas during my 4 years of studying it at university level and did splendid b/c of understanding derivations...

Answer 21

here is the proof @mashe

Answer 22

\[\sum_{k=1}^{k=n} k = n (n-1)/2\]