Home work with set builder notation

AI Thread Summary
The discussion focuses on converting the set (2, 5, 10, 17) into set builder notation, highlighting the challenge of finding a useful condition for such a small set. Participants note that the differences between terms suggest a pattern involving the addition of odd numbers. Several proposed set builder notations are deemed impractical or overly complex for the limited set. A more detailed approach involves expressing terms as a function of a summation, but there is uncertainty about the validity of the pattern beyond the initial four terms. Overall, the conversation emphasizes the difficulty of applying set builder notation to small, specific sets.
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Ok, I am needing help turning (2, 5, 10, 17) into set builder notation. I know to get these you add odd numbers 3, 5, 7 but I can't wrap my mind around putting this into notation.
 
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In other words, you are looking for a formula. Check for a changing difference between consecutive terms.
 
It's pretty pointless to use set builder notation for such a small set. (Homework always seems that way, doesn't it?)

Keep in mind that set builder notation is of the form

{expression | for <variable(s)> in {a bigger set} such that <condition>}

Here, you're working with integers, so the "bigger set" is going to be Z or Z+ or something.

The tricky part is figuring out a useful condition. For example, if your set was {2, 3, 5, 7, 11}, you could have said: {x | x in Z+ where x is prime and x <= 11}.
 
This question is silly. Here are some equally silly answers.

\{n | n\in\{2, 5, 10, 17\}\}
\{n | (n-2)(n-5)(n-10)(n-17)=0\}
\{n^2+1 | 1\le n\le4\}

The polynomial in the second answer can be rewritten as n^4 - 34n^3 + 369n^2 - 1460n + 1700, if you prefer.
 
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How exactly to convert this into set builder notation, not sure; but I did some checking on the sequence of numbers.

The first term is obviously just 2.
After that, the next terms conform to 2 plus the sumation as index goes from 2 to i of three plus two times the expression (n-2);

In other words, I'm saying from the second term onward, the term is
2 + summation from 2 to i of (3 + 2(n-2)).

Some variation from that pattern might be possible (not sure) after n=4, since we might not be sure if only four terms as originally given were enough to build the pattern.
 
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