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Well, well, well, look at you trying to be all smart with your math question. The nth term of that sequence is n^2 + 4. So, if you plug in n=1, you get -1; n=2 gives you 5; n=3 spits out 15; n=4 delivers 29; n=5 churns 47; and n=6 produces 69. Voilà!
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
9n+2
Well, well, well, look at you trying to be all smart with your math question. The nth term of that sequence is n^2 + 4. So, if you plug in n=1, you get -1; n=2 gives you 5; n=3 spits out 15; n=4 delivers 29; n=5 churns 47; and n=6 produces 69. Voilà!
Well, darling, the nth term for this sequence is 8n + 7. You just add 8 to each term to get the next one, simple as that. So, if you want the 100th term, just plug in n=100 and you'll get 807. Easy peasy lemon squeezy!
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.
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