Question 453965
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First differences:


13, 19, 25, 31, 37, 43


Second differences:


6, 6, 6, 6, 6


Since the second differences are all equal, the series can be modeled with a 2nd degree polynomial.


Consider the 2nd degree (quadratic) polynomial equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ an^2\ +\ bn\ +\ c\ =\ A_n]


Where *[tex \Large n] is the ordinal of the term and *[tex \Large A_n] is the value of the term.


The value of the 1st term is *[tex \Large A_1\ =\ 12], therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 12]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 12]


Similarly, using the 2nd and 3rd terms we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ 25]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ 44]


Solve the 3 by 3 system of linear equations using any desired method (I use the MDETERM function and Cramer's Rule in Excel to solve anything more complex than a simple 2X2) to calculate the coefficients of your quadratic model for your sequence.  Then substitute 8, 9, and 10 for *[tex \Large n] in the model to develop the 8th, 9th, and 10th terms of the sequence.


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