Not sure about "completing" the pattern since there following is the solution from an infinite sequence. You can fit the polynomial:
t(n) = (-45n6 + 993n5 - 8385n4 + 33895n3 - 66930n2 + 59192n - 15360)/240 with n = 1, 2, 3, ...
and the next term will be -308.
The 2 numbers in a row are multiplied, then you subtract one and that's the next number in the sequence. EG, 2x2 = 4 4-1=3 2x3=6 6-1=5 3x5=15 15-1=14 number sequence is 2, 2, 3, 5, 14
77-49=28÷2=〔14〕 21+14=35 35+14=49 49+14=63 63+14=77
There are not many numbers in the sequence here, but one solution that holds for this series is to take the first number, double it to get the second number, add 1 to get the next number, and repeat. So, the pattern would continue as follows: 1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63
14
8
25
The 2 numbers in a row are multiplied, then you subtract one and that's the next number in the sequence. EG, 2x2 = 4 4-1=3 2x3=6 6-1=5 3x5=15 15-1=14 number sequence is 2, 2, 3, 5, 14
77-49=28÷2=〔14〕 21+14=35 35+14=49 49+14=63 63+14=77
14
8
There are not many numbers in the sequence here, but one solution that holds for this series is to take the first number, double it to get the second number, add 1 to get the next number, and repeat. So, the pattern would continue as follows: 1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63
18
The pattern is +4, 2x, +4, 2x, and so on. The next number is 32.
It would be 14
The next number has to be 41; you always add the powers of three to the previous number beginning with 0. 3^0=1; 1+1=2; 3^1=3; 2+3=5; 3^2=9; 5+9=14; 3^3=27; 14+27=41 the next number would be 122 (81+41); you could also multiply the number by 3 and then substract 1. By the way: x(n)=x(n-1)+3^(x-2) ------------------------------------------------------------------------------------------------------------------- It could also be the Catalan number sequence, found by 1/(n+1) * (2n choose n), in which case the next number would be 42.
twenty five
Why would you want a baby at 14 you still have 3 or 4 year of school yet to complete.