1 4 9 is a series of squared numbers.
The nth term is [ n squared ]
tn = n2
7 - 4n where n denotes the nth term and n starting with 0
Un = (-1)n*(2n - 1)
It is increasing by 4 and the nth term is 4n+1
If you notice, there is a common difference between the terms: tn - tn-1 = -4 So the nth term is: tn = tn-1 - 4 For this recursive sequence to be defined though, you need something to start with as your tn-1. So start with t1= 3 and you're done.
The nth term is n2.
7n+4
n2
This is an Arithmetic Series/Sequence. In general the nth term, A(n) = a + (n - 1)d....where a is the 1st term and d is the common difference. In this question, the 1st term equals 1 and the common difference is 4. Then the nth term, A(n) = 1 + (n - 1) x 4 = 1 + 4n - 4 = 4n - 3.
Formula for nth termTn = a + (4n - 1) {where a is the first term and n is natural number}
tn = n2
n^2 + 2n + 1
4 Four Qutro The correct answer is: The nth term is 4n + 1
The nth term of the sequence is 2n + 1.
n-squared, or n to the power 2
n+4
I guess the first term should be -1 and not 1, then: The common difference of {-1, 4, 9, 14, 19, ...} is 5. Thus the nth term is given by 5n - 6 → 17th term is 5×17 - 6 = 79.