Question 278912
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The sum of the first 100 odd numbers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,{2i\ -\ 1}]


The sum of the first 100 numbers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,{i}]


The first 100 odd minus the first 100


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,\left(2i\ -\ 1\right)\ -\ \sum_{i=1}^{100}\,\left(i\right)\ =\ \sum_{i=1}^{100}\,\left(2i\ -\ 1\right)\ -\ \left(i\right)\ = \ \sum_{i=1}^{100}\,\left(i\ -\ 1\right)]


The first term of the resulting series (*[tex \Large i\ =\ 1]) is *[tex \Large 1\ -\ 1\ =\ 0]


The last term of the resulting series (*[tex \Large i\ =\ 100]) is *[tex \Large 100\ -\ 1\ =\ 99]


And there are still *[tex \Large n\ =\ 50] terms, so the sum is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(a\ +\ l\right)\left(n\right)}{2}\ =\ \frac{(0\ +\ 99)(100)}{2}\ =\ 99\,\cdot\,50\ =\ 4950]


Verification that  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,{2i\ -\ 1}\ =\ 10,000]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,{i}\ =\ 5050]


and therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{100}\,\left(2i\ -\ 1\right)\ -\ \sum_{i=1}^{100}(i)\ =\ 10000\ -\ 5050\ =\ 4950]


is left as an exercise for the student.


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