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This is what I have so far (similar/ the same as Pickett's solutions)

P: 7, 5, 3, 1

7-2(n-1)

Q: 1, 8, 27, 64

n3

R: 1/2, 2/3, 3/4, 4/5

n/(n+1)

S: 4, 9, 16, 25

(n+1)2

T: 1, 3, 9, 27

3(n-1)

U: 3, 6, 7, -2

8th Terms (n=8)

p = -7

q = 512

r = 8/9

s = 81

t = 2187

u = ?

Edited by David W
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I wrote it down on my calculator so unless you wrote the equation wrong your answer is wrong. I tried it on the first 3 terms.

Actually his equation is correct, but your calculator may be wrong. Be sure to make the calculations in the correct order: parentheses first, then powers, then multiplication and division, and finally addition/subtraction.

For n=1, we have n^3=n^2=n=1, so -(4/3)*1 + 7*1 - (26/3)*1 + 6 = -4/3 + 7 - 26/3 + 6 = -30/3 + 13 = -10 + 13 = 3

For n=2, we have n^3=8, n^2=4, so -(4/3)*8 + 7*4 - (26/3)*2 + 6 = -32/3 + 28 -52/3 + 6 = -84/3 + 34 = -28 + 34 = 6

For n=3, we have n^3=27, n^2=9, so -(4/3)*27 + 7*9 - (26/3)*3 + 6 = -108/3 + 63 - 78/3 + 6 = -186/3 + 69 = -62 + 69 = 7

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Technically speaking, for any series of n numbers, you can calculate an (n-1)-th degree polynomial that gets those answers through numerical methods, and there are an infinite number of higher degree polynomials that will get the same answer. (Recall that an nth degree polynomial can have n zero points. Likewise, it can be defined to have n + 1 "specified points" with the "zero" degree number acting as the first in the series).

Usually, I find these puzzles are also to be solved with the simplest solution possible (limiting the solution to the "best" answer).

Edited by Palustrius
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