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.
29 x 47 = 1363
29 and 47 are prime.
17, 19, 23, 29, 31, 37, 41, 43, and 47
17 19 23 29 31 37 41 43 47 53.
9n+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.
It is T(n) = n2 + 4*n + 2.
The given sequence is an arithmetic sequence with a common difference of 8. To find the nth term, we use the formula for the nth term of an arithmetic sequence: ( a_n = a_1 + (n-1)d ), where ( a_n ) is the nth term, ( a_1 ) is the first term, ( n ) is the position of the term, and ( d ) is the common difference. In this case, the first term ( a_1 = 15 ) and the common difference ( d = 8 ). Therefore, the nth term for this sequence is ( a_n = 15 + (n-1)8 = 15 + 8n - 8 = 8n + 7 ).
four hundred and nity six
81
To find the nth term of a sequence, we first need to find the pattern or rule that governs the sequence. By examining the differences between consecutive terms, we can see that the sequence is increasing by 6, 10, 14, 18, and so on. This means that the nth term is given by the formula n^2 + 4, where n represents the position of the term in the sequence.
1 3 4 7 11 18 29 47 76 123 199 .......and so on Add together the previous two terms to find the next term i.e 1+3=4, 3+4=7, 4+7=11
There are infinitely many possible answers. Given ANY number, it is always possible to find a polynomial of order 5 [at most] that can be used as the nth term rule for the given five number and the additional sixth. There are also non-polynomial solutions. Each different sixth number will result in a different polynomial and, since there are infinitely many sixth numbers, there are infinitely many answers to the question. Having said that, the simplest polynomial solution is Un = 9n - 2
One possible answer is t(n) = (n5 - 10n4 + 55n3 - 110n2 +364n)/60
29 x 47 = 1363
29