Question 186055
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  1 + 100 = 101]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2 + 99 = 101]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3 + 98 = 101]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  .]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  .]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  .]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  48 + 53 = 101]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  49 + 52 = 101]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  50 + 51 = 101]


So, figure out how many pairs you have that add to 101 and multiply times 101.  Hint: the number of pairs is half the number of numbers in your series.


In general:


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


Where <i>a</i> is the first number, <i>l</i> is the last number, and <i>n</i> is the number of numbers.  Note that there is no requirement that the numbers be consecutive.  For example, this would give you the sum of all the even numbers from 2 to 100 as long as you specify <i>a</i>, <i>l</i>, and <i>n</i> appropriately.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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