Question 597375
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,0}^4\,i^2]


The sum of the first *[tex \Large n] integers squared is given by *[tex \Large \frac{n(n\ +\ 1)(2n\ +\ 1)}{6}]


Substitute 4 for *[tex \Large n] and do the arithmetic.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,0}^4\,i^2\ =\ \frac{4(4\ +\ 1)(2(4)\ +\ 1)}{6}]


You get to do your own arithmetic.


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