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
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.
Sahara Desert
Havana, Cuba
aCalci Calcutta
Why does this site not give good answers? That i don't understand
Yes, that would be the geographical center of Mexico.
Latitude: N 26° 55' 3.8359", Longitude: W 80° 45' 56.8094"
27*23n+17*102n27*23=62117*102=1734621+1734=23552355nFactors of 2355n2355n/n=23552355/5=471471/3=1572355n=n*5*3*15711 is NOT a factor
2.9n + 1.7 = 3.5 + 2.3n Multiply each term by 10: 29n + 17 = 35 + 23n (gets rid of the decimals) Subtract 17 from both sides: 29n = 18 + 23n (gets rid of number from the left) Subtract 23n from both sides: 6n = 18 (gets rid of n from the right) Divide both sides by 6: n = 3 You do not need to carry out the first step if you are comfortable working with decimals.
15
-8. The numbers can be generated by the following quadratic: t(n) = -6n2 + 23n - 4 for n = 1, 2, 3, ...
The swift code and BIC code are the same for this bank it is NWBK GB 21 23N
That point is in far southern Algeria, about 37 miles northwest of the townof Tamanrasset, and 960 miles south-southeast of the capital at Algiers.