There are infinitely many rules based on polynomials of order 5 or above, as well as non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
The simplest rule, based on a polynomial or order 2 is
Un = n^2 - 1.
But it can also be
Un = (-n^5 + 15n^4 - 85n^3 + 245n^2 - 274n + 100)/20.
0,3,8,15,24,27,32,41
24 - 6n
This is an arithmetic progression. In general, If an A.P. has a first term 'a', and a common difference 'd' then the nth term is a + (n - 1)d. In the sequence shown in the question, the first term is 0 and the common difference is 5, therefore the nth term is, 0 + (n - 1)5. This can be rearranged to read : 5(n - 1) For example : the 7th term is 30 : 5(7 - 1) = 5 x 6 = 30.
You can see that you add 10 to the previous term to get the next term. Term number 1 2 3 4 Term 4 14 24 34 You can also say: Term number 1 2 3 4 Term 0*10+4 1*10+4 2*10+4 3*10+4 So the nth term would be 10(n-1)+4 Or if you expand it, it's 10n-6
An = 2(n - 1)2 + 2(n - 1) = 2n(n - 1)
0,3,8,15,24,27,32,41
The nth term of that series is (24 - 6n).
24 - 6n
0
This is an arithmetic progression. In general, If an A.P. has a first term 'a', and a common difference 'd' then the nth term is a + (n - 1)d. In the sequence shown in the question, the first term is 0 and the common difference is 5, therefore the nth term is, 0 + (n - 1)5. This can be rearranged to read : 5(n - 1) For example : the 7th term is 30 : 5(7 - 1) = 5 x 6 = 30.
n - 1
7 - 4n where n denotes the nth term and n starting with 0
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
n-9+3
You can see that you add 10 to the previous term to get the next term. Term number 1 2 3 4 Term 4 14 24 34 You can also say: Term number 1 2 3 4 Term 0*10+4 1*10+4 2*10+4 3*10+4 So the nth term would be 10(n-1)+4 Or if you expand it, it's 10n-6
An = 2(n - 1)2 + 2(n - 1) = 2n(n - 1)