One possible solution is the nth term is (-1)ⁿ for n = 1, 2, .., 5.
Beyond n = 5, we have no information as to how the sequence continues, The first 10 terms may be:
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The problem with a sequence is that it is the zeros of t{n} - a{n}.
The solution will be a polynomial of degree equal to the number of terms given.
However, there are infinitely many polynomials of a given degree that will go through the zeros given. There are some polynomials of a lesser degree which will go through the given zeros.
eg for a{n} = {1, 2, 3, 4, ...} the counting numbers, a polynomial of degree 1 will suffice: t{n} = n. However, it is possible to find a polynomial of degree 4 which will permit any value to follow;
eg for the given a{n}, t{n} = -(x⁴ - 10x³ +35x² - 62x + 24)/12 will also give a{n} = {1, 2, 3, 4} for n = 1, 2, 3, 4, but a{5} = 3; it has the same 4 first terms and then continues with a different 5th and subsequent terms.
According to Wittgenstein's Finite Rule Paradox every finite sequence of numbers can be a described in infinitely many ways and so can be continued any of these ways - some simple, some complicated but all equally valid.
The simplest polynomial rule here is T(n) = 5n2 - 15n - 1 for n = 1, 2, 3
It is: nth term = 35-9n
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
Clearly, if you omit the sign, the nth. term will be 4n. The alternating sign can easily be expressed as a power of (-1), so in summary, the nth. term is (-1)n4n.
-1
The nth term of the sequence is 2n + 1.
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.
The given sequence is an arithmetic sequence with a common difference of 6. To find the nth term of this sequence, we can use the following formula: nth term = first term + (n - 1) x common difference where n is the position of the term we want to find. In this sequence, the first term is 1 and the common difference is 6. Substituting these values into the formula, we get: nth term = 1 + (n - 1) x 6 nth term = 1 + 6n - 6 nth term = 6n - 5 Therefore, the nth term of the sequence 1, 7, 13, 19 is given by the formula 6n - 5.
It is: nth term = 29-7n
It is: nth term = 35-9n
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
The nth term is: 5-6n
12 - 5(n-1)
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
The nth term of the sequence is 3n - 2.
123456789 * * * * * The nth term is 3n
The nth term is 4n-1 and so the next term will be 19