Question 991254
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The *[tex \Large n]th term of the first sequence is *[tex \Large n^2]


The *[tex \Large n]th term of the second sequence is the *[tex \Large n-1]th term plus *[tex \Large n].  That is, the first term is 0 + 1 = 1, the second term is 1 + 2 = 3, the third term is 3 + 3 = 6, and so on...


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \