tn = n2
tn = 34 - 9n where n = 1,2,3,...
The given sequence is an arithmetic sequence where each term increases by 4. The first term (a) is 13, and the common difference (d) is 4. The nth term can be found using the formula: ( a_n = a + (n-1)d ). Therefore, the nth term is ( a_n = 13 + (n-1) \cdot 4 = 4n + 9 ).
The sequence given consists of the squares of the natural numbers: (1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2). To find the nth term of the sequence, you can use the formula (n^2), where (n) is the position in the sequence. Therefore, the nth term is (n^2).
The nth term is 7n-3 and so the next term will be 39
7n - 3
Formula for nth termTn = a + (4n - 1) {where a is the first term and n is natural number}
The nth term is: 5-6n
The nth term is 9n-2
It is 4n+5 and so the next term will be 25
t(n) = 28-3n where n = 1,2,3,...
n2
The nth term = 9n-2
(n+1)^2 Please tell me you know what that means.
Sn = 3n2 + 2n - 8
tn = 34 - 9n where n = 1,2,3,...
n-squared, or n to the power 2
As given, the sequence is too short to establish the generating rule. If the second term was 19 and NOT 29, then the nth term is tn = 6*n + 7 or 6(n+1)+1