There are infinitely many polynomials of order 6 (or higher) that will give these as the first six numbers and any one of these could be "the" rule. 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.
In this particular case, the simplest solution isU(n) = 3*n^2 - 1 for n = 1, 2, 3, ...
It is: nth term = 7n-9
To find the nth term in a quadratic sequence, we first need to determine the pattern. In this case, the difference between consecutive terms is increasing by 3, 5, 7, 9, and so on. This indicates a quadratic sequence. To find the 9th term, we need to use the formula for the nth term of a quadratic sequence, which is given by: Tn = an^2 + bn + c. By plugging in n=9 and solving for the 9th term, we can find that the 9th term in this quadratic sequence is 74.
foot
(13 + 15 + 26 + 11 + 26 + 16 + 12)/7 = 17 ====
n2 + 3n - 2
To find the nth term of this sequence, we first need to determine the pattern or rule governing the sequence. By examining the differences between consecutive terms, we can see that the sequence is increasing by 9, 15, 21, 27, and so on. This indicates that the nth term is given by the formula n^2 + 1.
The nth term in this arithmetic sequence is an=26+(n-1)(-8).
It is: nth term = 7n-9
It is: nth term = 35-9n
46n9
It is: 26-6n
[ 6n + 8 ] is.
Tn = 10 + n2
To find the nth term of the sequence 11, 21, 35, 53, 75, 101, we can observe the differences between consecutive terms: 10, 14, 18, 22, and 26, which increase by 4 each time. This suggests that the sequence can be described by a quadratic function. The nth term can be represented as ( a_n = 5n^2 + 6n ), where n starts from 1. Thus, the nth term corresponds to this formula for values of n.
Well, isn't that just a lovely pattern we have here? Each term is increasing by 4, isn't that delightful? So, if we want to find the nth term, we can use the formula: nth term = first term + (n-1) * common difference. Just like painting a happy little tree, we can plug in the values and find the nth term with ease.
The common difference (d) between successive terms is -9. The first term (a) is 26 The formula for the nth term [a(n)] of an Arithmetic Series is , a + (n - 1)d. Inputting the values for a and d gives :- a(n) = 26 - 9(n - 1) = 26 - 9n + 9 = 35 - 9n......where n = 1,2,3......
t(n) = 4n2 - 4n + 2