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
The sequence given is 8, 14, 22. To find the pattern, we can observe that the differences between the numbers are 6 (14 - 8) and 8 (22 - 14). Following this pattern, the next difference might increase by 2, suggesting the next number is 22 + 10 = 32. Continuing this pattern, the subsequent number would be 32 + 12 = 44. Therefore, the next two numbers are 32 and 44.
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
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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
When you subtract 14 from a number, you are essentially reducing that number by 14. The result can be expressed as "the number minus 14." For example, if the number is represented by "x," the expression would be "x - 14." The outcome will vary depending on the value of the original number.
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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.
Why would you want a baby at 14 you still have 3 or 4 year of school yet to complete.
14 = fourteen
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