Question 187230
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sum_{i=1}^{25}(4i + 1)]


This means to replace <i>i</i> in 4<i>i</i> + 1 with the integers from 1 to 25 and add up the results.  So the first term is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(1) + 1 = 5]


and the second term is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(2) + 1 = 9]


and so on until you get to the last term which is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(25) + 1 = 101]


Now you could do this the hard way, namely directly compute the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5 + 9 ... 97 + 101]


But that method is long, arduous, and prone to error.  Fortunately, there is a formula for the sum of any series of numbers:


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


Where <i>a</i> is the first element of the series, <i>l</i> is the last element of the series, and <i>n</i> is the number of elements.  So for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sum_{i=1}^{25}(4i + 1) =\ \frac{(5 + 101)25}{2}]


You can do your own arithmetic.



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