Question 1192881:  What is the 20th term of the sequence 0, 3, 8, 15, 24, 35, 48, . . 
 
 Found 2 solutions by  josgarithmetic, math_tutor2020: Answer by josgarithmetic(39630)      (Show Source):  Answer by math_tutor2020(3817)      (Show Source): 
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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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