1 -1, 2 -2, 3 -3, 4 -4, and so on. It is simply a positive number followed by its negative number.
You add the 2 numbers before e.g. 2+3=5
The answer is (no.) x3 = (no. 2) - 1 = (no. 3) x3= (no.4) x4 and so on... 1 x 3 = 3 - 1 = 2 x 3 = 6 - 1 = 5 x 3 = 15 - 1 = 14 x 3 = 42 - 1 = 41. :D Hope i helped
12 ****************************** 11, 16, 22 are the 3 numbers continuing the pattern.
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
plus 4, minus 3
Well, isn't that just a happy little pattern we have here? It looks like the pattern rule is adding 3, then adding 1, then adding 2, and repeating. So, we start at 0, add 3 to get 3, add 1 to get 4, then add 2 to get 6, and the pattern continues from there. Just remember, there are no mistakes in patterns, only happy little accidents.
t(n) = 3(n-1) + 1, for n = 1, 2, 3, etc
-11 Pattern: Subtract 1, subtract 2, subtract 3 and so on.
The numbers are what you get when you make a sum of reciprocal exponents. N(1) = 1^1 = 1 N(2) = 1^2 + 2^1 = 1 + 2 = 3 N(3) = 1^3 + 2^2 + 3^1 = 1 + 4 + 3 = 8 N(4) = 1^4 + 2^3 + 3^2 + 4^1 = 1 + 8 + 9 + 4 = 22 The next number in the pattern would be 2780.
1:2:3:5:8:5:3:2:1
It is 2, assuming the pattern is repeated as given. 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1... If the intended pattern is to continue to subtract 1 from the last number, then the 5479th digit of the pattern will be -5470.
Start at 1. Multiply by 3. Subtract 1. Multiply by 3. Subtract 1. Repeat this pattern.
Add the previous 2 numbers to get the next number.
1123581358 1+1=2 1+2=3 3+5=8 5+8=13 3+5=8
this series increments in powers of 3 like 1+3^0=2 2+3^1=5 5+3^2=14 14+3^3=41 and so on....
You start with two integers, usually 0 and 1 or 1 and 1 and get the next number in the pattern by adding the two previous numbers. For example starting with 1 and 1 , 1 + 1 = 2 ,so the sequence is now 1 1 2 1+ 2 = 3 , so the sequence is 1 1 2 3 2+3 = 5 . " " " " 1 1 2 3 5 The formula is an = an-1 + an-2. where a0 and a1 are given and n > 1.
You add the 2 numbers before e.g. 2+3=5