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What is the nth term and sum of n terms of 6 13 24 39?
The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6
What is the next number 8 6 9 10 4 11?
Any number can be the next number. It is easy to find a rule based on a polynomial of order 6 such that the first six numbers are as listed in the question followed by the chosen next number. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule, based on a polynomial of order 5, isT(n) = (23n^5 - 335n^4 + 1715n^3 - 3685n^2 + 3122n + 120)/120 where n = 1, 2, 3, ...Accordingly, the next number is T(7) = 99.
