It is probably 2n.
But there are infinitely many polynomials of degree 5, as well as other mathematical functions, that will fit the above 4 points.
For example, one possibility for the nth term is
Un = (n4 - 10n3 + 35n2 - 42n - 24)/4
The nth term is (2n - 12).
If you mean: 3, 4, 5, 6 and 7 then nth term = n+2
It is: nth term = -4n+14
The nth term is 2n So the 20th term is 2 x 20 = 40.
1 2 3 4= n 2 4 6 8 plusing two = 2n answer 2n
The nth term is (2n - 12).
If you mean: 3, 4, 5, 6 and 7 then nth term = n+2
The sequence 4, 6, 8, 10 is an arithmetic sequence where each term increases by 2. The nth term formula can be expressed as ( a_n = 4 + (n - 1) \cdot 2 ). Simplifying this gives ( a_n = 2n + 2 ). Thus, the nth term of the sequence is ( 2n + 2 ).
It is: nth term = -4n+14
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
The nth term is 2n So the 20th term is 2 x 20 = 40.
1 2 3 4= n 2 4 6 8 plusing two = 2n answer 2n
The given sequence 6, 8, 10, 12 is an arithmetic sequence with a common difference of 2 between each term. To find the nth term of an arithmetic sequence, you can use the formula: (a_n = a_1 + (n-1)d), where (a_n) is the nth term, (a_1) is the first term, (n) is the term number, and (d) is the common difference. In this case, the first term (a_1) is 6 and the common difference (d) is 2. So, the nth term (a_n = 6 + (n-1)2 = 2n + 4).
The nth term of the sequence is (n + 1)2 + 2.
To find the nth term of the sequence 4, 13, 28, 49, 76, we first identify the differences between consecutive terms: 9, 15, 21, 27. The second differences, which are constant at 6 (6, 6, 6), suggest that the sequence is quadratic. The nth term can be expressed as ( an^2 + bn + c ). By solving the equations based on the first few terms, we find the nth term is ( n^2 + 3n ).
The nth term is 9n-2
Un = 2n + 2 is one possible answer.