Question 251892
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Don't know where you went wrong with part B, but if you think about it for a second, you know that the answer cannot have a decimal fraction.  You are adding a series of integers, and the integers are closed for addition.  That means any time you add two integers, you get another integer.


Part B:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n(n\,+\,1)(2n\,+\,1)}{6}\ =\ \frac{30\,\cdot\,31\,\cdot\,61}{6}]


Try that arithmetic one more time.


Part C:


You have the correct answer.


Part D and E


First calculate the sum of the integers from 1 to 30:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n(n\,+\,1)}{2}\ = \frac{30\,\cdot\,31}{2}]


Square the result for part D and cube the result for part E.


That is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ D = \left(\frac{30\,\cdot\,31}{2}\right)^2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ E = \left(\frac{30\,\cdot\,31}{2}\right)^3]


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