What does 1-2+3...-98+99-100 make

You can write this as follows
(1+3+5+.............99) - ( 2+4+6+..............+98+100)

So now here we have two arithmetic The sum of the first odd numbers up to and including 99 minus the sum of the first even numbers up to and including 100. Eacg series has 50 terms.

Now, the sum of an arithmetic series with initial term a, and last terk L is given by
s=(a+L)*n/2

Using this the sum of the series in the question is therefore

S - (1 + 99)*50/2 + (2 +100)*50/2 =100*25-102*25=25*(100-102) = -50

100 steps performed completely = -50

Effectively each step moves an increment of 1 in either direction, so 100 steps will have gone positive and negative in turn,
...& knowing it results in -100 applied to a positive from just the first 1-2+3[=2]-4=-2 stages means you can reasonably assume that [taking a shortcut] at -10 it will result in -5, at -50 it will give -25, so -100 will equal -50

Is it the same for 1+2-3...+98-99+100 ?
1+2-3+4=4
...from this I'd assume the answer is instead 100


What do you make of this though;
http://christianhilton.xanga.com/6395463…

It's the same kind of premise, although it's applied to a denominator described as 1/[n-1]
If you take away 1/4 of the whole [100] from the whole, you add 1/3 of the result to the result
So,
100 - 25 = 75, (1/4 of 100 is 25)
75 + 25 = 100 (1/3 of 75 is 25)
For this reason I call the formula using 1/[n-1], "Christian-1's Constant" until otherwise reliably informed...

1-2+3...-98+99-100 = 1+3+5...+99 -(2+4+6....+100)

1-2+3...-98+99-100 = (50/2)(1+99) - (50/2)(2+100)

1-2+3...-98+99-100 = 2500 - 2550

1-2+3...-98+99-100 = -50 >============< ANSWER

∑_(〖-1〗^(n+1))×n〗

n: from 1 until 100
∑_=multiple of variable

1-2=-1
3-4=-1
...
99-100=-1

50 times so it equals -50

well 50 must be the answer as above, but how come i got -97 from the Calculator?

(1-2)×100÷2=-50