There are infinitely many possible answers.
The only solution in the form a cubic polynomial is
Un = (25n3 - 147n2 + 236n - 120)/12 for n = 1, 2, 3, 4
Each term is 6 more than the last term; ie add 6 to the current term to get the next term. U{n} = 6n + 4 for n = 1, 2, 3, 4
The rule is multiply the previous term by -1 to find the next term.
Each term is increased by adding five times three to an increasing exponent. 2 + 5 x 3 to the zero = 7 + 5 x 3 to the first = 22 + 5 x 3 squared = 67 The next term would be 202 which makes the rule something like n = n + 5 x 3 to the x+1, but please double-check that.
In a Geometric Sequence each term is found by multiplying the previous term by a common ratio except the first term and the general rule is ar^(n-1) whereas a is the first term, r is the common ratio and (n-1) is term number minus 1
0.5n(n+1)
Term to Term rule in Maths is how much you go up or down in. e.g 1,2,3,4,5,6 would be +1
VH-1 Top 20 Video Countdown - 1994 2004-05-22 was released on: USA: 22 May 2004
Each term is 6 more than the last term; ie add 6 to the current term to get the next term. U{n} = 6n + 4 for n = 1, 2, 3, 4
The rule is multiply the previous term by -1 to find the next term.
Each term is increased by adding five times three to an increasing exponent. 2 + 5 x 3 to the zero = 7 + 5 x 3 to the first = 22 + 5 x 3 squared = 67 The next term would be 202 which makes the rule something like n = n + 5 x 3 to the x+1, but please double-check that.
It is increasing by odd numbers consecutively._______or: n = the term number, the rule is: n2 or n(n). 12=1(1x1=1), 22=4(2x2=4), 32=9(3x3=9), and so on.
PRIME
In a Geometric Sequence each term is found by multiplying the previous term by a common ratio except the first term and the general rule is ar^(n-1) whereas a is the first term, r is the common ratio and (n-1) is term number minus 1
22 is an integer, not a fraction!
0.5n(n+1)
The nth term is 4n - 3
Rule 1: The term is integer, not interger.Rule 2: The answer depends on what you want to do with it or them.