Question 1012259
The tetrahedral numbers are found in Pascal's triangle, namely
Tn = combination of (n+2) taken 3 at a time = (n+2)! / (3! * (n+2-3)!) =
(n+2)! / (3! * (n-1)!) where Tn is the nth tetrahedral number
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another way to solve this is to note that the sum of the first n triangular numbers is the nth tetrahedral number,that is
Tn = (n(n+1)(n+2)) / 6
a triangular number can be thought of as the number of dots needed to form an equilateral triangle, namely
1, 3, 6, 10, 15, 21, and so forth
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the nth sum(Sn) of n consecutive tetrahedral numbers is
Sn = (n(n+2)(n+3)) / 24
We can derive this formula by starting with
Tn = (n(n+1)(n+2)) / 6, then expand the numerator
Tn = (n^3 + 3n^2 + 2n) / 6
Next we use the formulas for the sum of first n consecutive integers, first n squared integers and first n cubed integers, namely
Sn = (n(n+1)) / 2
Sn^2 = (n(n+1)(2n+1)) / 6
Sn^3 = (n^2(n+1)^2 / 4
Substitute these equations into our Tn formula and simplify 
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