For the given sequence the simplest formula for the nth term is:
U{n} = n² + 3n - 2
However, that will only work if the sequence continues 68, 86, 106, 128, 152, ... (the difference between successive terms being the difference between the previous 2 terms plus 2).
If the sequence continues differently, then a different polynomial formula with higher powers will be required.
Any number that you choose can be the nth term. 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 in any of these ways - some simple, some complicated but all equally valid. 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.
Using the principle of Occam's razor, the simplest solution, based on a polynomial of degree 2,
U(n) = n^2 + 3n - 2
2 + ((6 + 2 * (n - 1) * (n - 1))
[ 6n + 8 ] is.
Each number is increasing by increments of 8 10 12 14 ... etc and so the next number will be 52+16 = 68
It depends what the next number in the sequence is. The simplest polynomial for those 5 terms is: U{n} = n² + 3n - 2
The nth term of the sequence is (n + 1)2 + 2.
2 + ((6 + 2 * (n - 1) * (n - 1))
[ 6n + 8 ] is.
Each number is increasing by increments of 8 10 12 14 ... etc and so the next number will be 52+16 = 68
It depends what the next number in the sequence is. The simplest polynomial for those 5 terms is: U{n} = n² + 3n - 2
The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.
The nth term of the sequence is (n + 1)2 + 2.
9n+2
n2 + 3n - 2
The 'n'th term is [ 13 + 5n ].
The 'n'th term is [ 13 + 5n ].
The 'n'th term is [ 13 + 5n ].
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