2/3n3
Given any number, it is always possible to find a polynomial of degree 6 that will fit the above numbers and the additional given number.The simplest position to value rule, in polynomial form, for the above sequence isUn = (3n3 - 5n2 + 4n - 12)/2 for n = 1, 2, 3, ...and accordingly, U7 = 412.
One possible answer is: Un = (3n3 - 3n2 + 78n - 8)/560 for n = 1, 2, 3, 4.
Given ANY number, it is easy to find a polynomial of order 4 such that if you use it as a position to value rule you get the four given numbers and your chosen one as the fifth. As a result, any number can be the next in the sequence. The simplest polynomial of order 3 for the above four numbers is: Un = (-3n3 + 32n2 - 57n + 34)/2 for n = 1, 2, 3, ... Accordingly, the next number is 87.
Suppose the smallest of the integers is n. Then the product of the four consecutive integers is n*(n+1)*(n+2)*(n+3) =(n2+3n)(n2+3n+2) = n4+6n3+11n2+6n So product +1 = n4+6n3+11n2+6n+1 which can be factorised as follows: n4+3n3+n2 +3n3+9n2+3n + n2+3n+1 =[n2+3n+1]2 Thus, one more that the product of four consecutive integers is a perfect square.
6590. One possible sequence is 2472/3N3 - 14421/2N2 + 26165/6N - 1390
2/3n3
2/3n3
The GCF is 3n3
Given any number, it is always possible to find a polynomial of degree 6 that will fit the above numbers and the additional given number.The simplest position to value rule, in polynomial form, for the above sequence isUn = (3n3 - 5n2 + 4n - 12)/2 for n = 1, 2, 3, ...and accordingly, U7 = 412.
One possible answer is: Un = (3n3 - 3n2 + 78n - 8)/560 for n = 1, 2, 3, 4.
Given ANY number, it is easy to find a polynomial of order 4 such that if you use it as a position to value rule you get the four given numbers and your chosen one as the fifth. As a result, any number can be the next in the sequence. The simplest polynomial of order 3 for the above four numbers is: Un = (-3n3 + 32n2 - 57n + 34)/2 for n = 1, 2, 3, ... Accordingly, the next number is 87.
Suppose the smallest of the integers is n. Then the product of the four consecutive integers is n*(n+1)*(n+2)*(n+3) =(n2+3n)(n2+3n+2) = n4+6n3+11n2+6n So product +1 = n4+6n3+11n2+6n+1 which can be factorised as follows: n4+3n3+n2 +3n3+9n2+3n + n2+3n+1 =[n2+3n+1]2 Thus, one more that the product of four consecutive integers is a perfect square.
The answer depends on which number is missing. Even if the location of the gap is know, there are infinitely many possible solutions. The solutions listed below are polynomials of the lowest degree. First: 5 using the rule Un = (-4n3 + 48n2 - 128n + 99)/3 Second: 0 using the rule Un = (-7n3 + 84n2 - 209n + 138)/6 Third: 21.666 (recurring) using the rule Un = (3n3 - 23n2 + 84n - 61)/3 Fourth: 50 using the rule Un = (-11n3 + 90n2 - 121n + 48)/6 Fifth: 65 using the rule Un = (-4n3 + 36n2 - 44n + 15)/3
Any number of your choice. It is possible to find a quartic (order 4) polynomial that will fit the given 4 points and any other. There is only one cubic that will do the trick: Un = (-3n3 + 26n2 - 63n + 52)/2 for n = 1, 2, 3, ... and according to it U5 = 6.
A solution for 4th difference sequence is:Formula: Tn = an4+ bn3+ cn2 + dn + e1st term = a + b + c + d + e1st difference = 15a + 7b + 3c + d2nd difference = 50a + 12b + 2c3rd difference = 60a + 6b4th difference = 24aExample:Tn (term number) = 1 2 3 4 5 6 7Sequence = 1 41 209 643 1529 3101 56411st difference = 40 168 434 886 1572 25402nd difference = 128 266 452 686 9683rd difference = 138 186 234 2824th difference = 48 48 48Step 1: 4th difference = 24a:24a = 48a = 2Step 2: 3rd difference = 60a + 6b:60(2) + 6b = 138b = 3Step 3: 2nd difference = 50a + 12b + 2c:50(2) + 12(3) + 2c = 128c = -4Step 4: 1st difference = 15a + 7b + 3c + d:15(2) + 7(3) + 3(-4) + d = 40d = 1Step 5: 1st 1st term = a + b + c + d + e:2 + 3 - 4 + 1 + e = 1e = -1Answer:Tn = 2n4+ 3n3- 4n2 + n - 1
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