Question 1192881
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I'm assuming that this is a quadratic sequence because the gap between terms 3, 5, 7, 9, 11, 13, ... are of an arithmetic sequence (common difference d = 2). 
If this assumption is not the case, then please let me know and I'll update the solution.


If it is quadratic, then it means you'll have something of the form
y = ax^2+bx+c
where x is the term number and y is the term itself.


Examples: 
(x,y) = (1,0) means the first term is 0.
(x,y) = (2,3) means the second term is 3.
and so on.


Plug in (x,y) = (1,0) to find that
y = ax^2+bx+c
0 = a(1)^2+b(1)+c
0 = a+b+c
a+b+c = 0


Then plug in x = 2 and y = 3 to get
4a+2b+c = 3


And plug in x = 3 and y = 8 to get
9a+3b+c = 8


The task is to solve this system of equations
a+b+c = 0
4a+2b+c = 3
9a+3b+c = 8
I'll skip the steps because it's a very long process, but the punchline is that
a = 1
b = 0
c = -1


This means
y = ax^2+bx+c
y = 1x^2+0x+(-1)
y = x^2-1
represents a nice simple formula to generate any term (y) based on the term number (x).


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Checking the formula:


Plug in x = 1 to represent the first term
y = x^2 - 1
y = 1^2 - 1
y = 0
That works out because 0 is indeed the first term


Repeat for x = 2
y = x^2 - 1
y = 2^2 - 1
y = 3
That works out as well


I'll let you check the other terms.


Plugging in x = 20 will then lead you to y = 399 which is the 20th term.


Once again, this all depends on if the sequence given is quadratic in nature.
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