Question 674615
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Use the method of common differences:


<pre>


      6     6     2    -6   -18
       \   / \   / \   / \   / 
         0    -4    -8   -12 
          \   / \   / \   /
           -4    -4    -4
</pre>


The differences become common at the 2nd level down, hence the desired model for the nth term is going to be a 2nd degree polynomial.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T(n)\ =\ an^2\ +\ bn\ +\ c]


The value of the term when n = 1 is 6, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 6]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 6]


Using similar logic, we develop:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ 6]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ 2]


Solve the 3X3 system to find the values of the coefficients for *[tex \LARGE T(n)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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