an^3 bn^2+cn+d=E(n) where E(n) is the value of the nth term and a,b, c and d are constants. So we have:
a+b+c+d=1
8a+4b+2c+d=0
27a+9b+3c+d=3
64a+16b+4c+d=8
Then work it out from there.
Assuming the first term was supposed to be -1 (minus 1) not 1 (plus one), then the nth term is given by:
tn = n2 - 2n
The nth term of the sequence is 2n + 1.
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.
To find the nth term of a sequence, we first need to identify the pattern or rule that governs the sequence. In this case, the sequence is decreasing by 6 each time. Therefore, the nth term can be represented by the formula: 18 - 6(n-1), where n is the position of the term in the sequence.
The nth term is: 5-6n
12 - 5(n-1)
The nth term of the sequence is 2n + 1.
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.
To find the nth term of a sequence, we first need to identify the pattern or rule that governs the sequence. In this case, the sequence is decreasing by 6 each time. Therefore, the nth term can be represented by the formula: 18 - 6(n-1), where n is the position of the term in the sequence.
The given sequence is an arithmetic sequence with a common difference of 6. To find the nth term of this sequence, we can use the following formula: nth term = first term + (n - 1) x common difference where n is the position of the term we want to find. In this sequence, the first term is 1 and the common difference is 6. Substituting these values into the formula, we get: nth term = 1 + (n - 1) x 6 nth term = 1 + 6n - 6 nth term = 6n - 5 Therefore, the nth term of the sequence 1, 7, 13, 19 is given by the formula 6n - 5.
7 - 4n where n denotes the nth term and n starting with 0
It is: nth term = 29-7n
It is: nth term = 35-9n
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
The nth term is: 5-6n
12 - 5(n-1)
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1