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.
plus 4, minus 3
0, 1, 1 (0+1), 2 (1+1), 3 (2+1), 5 (2+3), 8 (3+5), 13(5+8), 21 (13+8), 34 (21+13),....and so on.
t(n) = 3(n-1) + 1, for n = 1, 2, 3, etc
Start at 1. Multiply by 3. Subtract 1. Multiply by 3. Subtract 1. Repeat this pattern.
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.
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
11
12 ****************************** 11, 16, 22 are the 3 numbers continuing the pattern.
yn = 1/2n2 + 2 1/2n + 3 y9 = 1/2(9)2 + 2 1/2(9) + 3 y9 = 40 1/2 + 22 1/2 + 3 y9 = 66
plus 4, minus 3
minus 6, plus 3, minus 6, plus 3...
Seperate like this: 2 1 2 3 2 6 2 10 2 ? then categorize them two by two: 2 1 2 3 2 6 2 10 2 ? now this is the question : 1 3 6 10 ? and the answer is 15" 1 + 2 = 3 ---> + 3 = 6 ----> + 4 = 10 ----> + 5 =15
0, 1, 1 (0+1), 2 (1+1), 3 (2+1), 5 (2+3), 8 (3+5), 13(5+8), 21 (13+8), 34 (21+13),....and so on.