The answer will depend on where, in the sequence, the missing number is meant to be. Also, in each case, there are infinitely many possible answers since it is possible to find a polynomial of degree 5 (or higher) that will go through each of the above numbers and ANY other number, missing from ANY position.
Using polynomials of order 4, though, there is only one answer for each position. For example,
First number missing: 171
Un= (-27n4+ 422n3- 2364n2+ 5611n - 4668)/6
Last number missing: 218
Un= (-27n4+ 314n3- 1260n2+ 2041n - 1026)/6
The answer will depend on where, in the sequence, the missing number is meant to be. Also, in each case, there are infinitely many possible answers since it is possible to find a polynomial of degree 5 (or higher) that will go through each of the above numbers and ANY other number, missing from ANY position.
Using polynomials of order 4, though, there is only one answer for each position. For example,
First number missing: 171
Un= (-27n4+ 422n3- 2364n2+ 5611n - 4668)/6
Last number missing: 218
Un= (-27n4+ 314n3- 1260n2+ 2041n - 1026)/6
The answer will depend on where, in the sequence, the missing number is meant to be. Also, in each case, there are infinitely many possible answers since it is possible to find a polynomial of degree 5 (or higher) that will go through each of the above numbers and ANY other number, missing from ANY position.
Using polynomials of order 4, though, there is only one answer for each position. For example,
First number missing: 171
Un= (-27n4+ 422n3- 2364n2+ 5611n - 4668)/6
Last number missing: 218
Un= (-27n4+ 314n3- 1260n2+ 2041n - 1026)/6
The answer will depend on where, in the sequence, the missing number is meant to be. Also, in each case, there are infinitely many possible answers since it is possible to find a polynomial of degree 5 (or higher) that will go through each of the above numbers and ANY other number, missing from ANY position.
Using polynomials of order 4, though, there is only one answer for each position. For example,
First number missing: 171
Un= (-27n4+ 422n3- 2364n2+ 5611n - 4668)/6
Last number missing: 218
Un= (-27n4+ 314n3- 1260n2+ 2041n - 1026)/6
The answer will depend on where, in the sequence, the missing number is meant to be. Also, in each case, there are infinitely many possible answers since it is possible to find a polynomial of degree 5 (or higher) that will go through each of the above numbers and ANY other number, missing from ANY position.
Using polynomials of order 4, though, there is only one answer for each position. For example,
First number missing: 171
Un= (-27n4+ 422n3- 2364n2+ 5611n - 4668)/6
Last number missing: 218
Un= (-27n4+ 314n3- 1260n2+ 2041n - 1026)/6
The missing number is 18
I think your last number should be 216 as the others are the cubes of numbers 1, 2, 3 and 5. The missing number is 4 cubed, ie 64.
456445
The answer depends on where, within the sequence, the missing number should have been.
64
If a progression goes 15 __ 21, then the missing number will dictate how the pattern increases with every number. If the missing number is 18, this means that the numbers go up in threes. The pattern would continue 24, 27, 30, 33...
The missing number is 18
I think your last number should be 216 as the others are the cubes of numbers 1, 2, 3 and 5. The missing number is 4 cubed, ie 64.
456445
33
The answer depends on where, within the sequence, the missing number should have been.
64
44 = 256
64 is missing; the numbers are cubes of 1,2,3,4,5,& 6.
The next number in this sequence should be 33.
The only number in that list which is not a square number is 27. 27 is a cube number.
64 should be in between 27 and 125.