t(n) = 3(n-1) + 1, for n = 1, 2, 3, etc
+2, +3, +2
Add 3n-1. For example 1 + 30 = 2 2 + 31 = 5 5 + 32 = 14 14 + 33 = 41
The gaps are 5, 10, 15, 20, ..... The rule becomes add 5(n-1) to the previous number So that the second number is found by adding 5 times (2-1) to 1 = 5+1 = 6 and the third number is found by adding 5 times (3-1) to 6 = 10+6 = 16 and the fourth number is found by adding 5 times (4-1) to 16 = 15+15 = 31 and so on....
There are many possible answers. One possible rule is: Un = (3n4 - 18n3 + 69n2 - 54n + 120)/8 for n 1, 2, 3, ...
2 1 2
Every number is double the previous number, where the first number is 1.
Add the increasing numbers by one - as in 0 1 2 3 4 5 6 etc 1 + 0 = 1 1 + 1 = 2 2 + 2 = 4 4 + 3 = 7 7 + 4 = 11 11 + 5 = 16 16 + 6 = 22
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.
One rule for this pattern is to add twice the previous value added 4 + 1 = 5 5 + 2×1 = 5 + 2 = 7 7 + 2×2 = 7 + 4 = 11 11 + 2×4 = 11 + 8 = 19 Continuing the next numbers would be: 19 + 2×8 = 19 + 16 = 35 35 + 2×16 = 35 + 32 = 67 ...
Add the previous 2 numbers to get the next number.
Everytime you move to the next number you add 1, then 2, then 3, etc.1+1=22+2=44+3=77+4=1111+5=1616+6=2222+7=2929+8=37
t(n) = 3(n-1) + 1, for n = 1, 2, 3, etc
The rule of this pattern is -2 + 6 +4 so the next number would be 16.
i0 = 4; in = in-1 - 3
Divide by 2 to get the next term.
multiplication pattern