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
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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
The sequence appears to be decreasing by 3, then increasing by 3. Following this pattern, starting from 33, if we subtract 3, we get 30. Thus, the missing number is 30. The complete sequence would be 33, 30, 24, 27.
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.
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The answer depends on where, within the sequence, the missing number should have been.