Question 447576
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500 + 1 = 501


499 + 2 = 501


498 + 3 = 501


. . .


251 + 250 = 501


And there are 500/2 = 250 such pairs.


In general, the sum of consecutive positive integers is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,a}^l\,(i)\ =\ \frac{(a\,+\,l)(l\,-\,a\,+\,1)}{2}]


So for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,1}^{500}\,(i)\ =\ \frac{(1\,+\,500)(500\,-\,1\,+\,1)}{2}\ =\ 501\ \times\ 250]


Super Double Plus Extra Credit:


What is the sum of the <i><b>even</b></i> integers between 200 and 300 inclusive?




John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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