Find a polynomial f(n) such that for all integers n>=1, we have
1*3 + 2*4 + 3*5 + ... + n*(n + 2) = f(n).
Write f(n) as a polynomial with terms in descending order or n.
+130071 Successive sums produced
3 11 26 50 85
8 15 24 35 [ successive differences ]
7 9 11
2 2
0
Sum of differences method
We have 3 rows of non-zero differences
This will produce a 3rd power polynomial an^2 + bn^2 + cn + d
We have the equations
a(1)^3 + b(1)^2 + c(1) + d = 3
a(2)^3 + b(2)^2 + c(2) + d = 11
a(3)^3 + b(3)^2 + c(3) + d = 26
a(4)^3 + b(4)^2 + c(4) + d = 50
Simlifying we have
a + b + c + d = 3
8a + 4b + 2c + d = 11
27a +9b + 3c + d = 26
64a + 16b + 4c + d = 50
This system is a little lengthy to solve (but not impossible !!! ) so I'll use a matrix solver
a = 1/3
b = (3/2)
c = (7/6)
d = 0
Our generating polynomial is
(1/3)n^3 + (3/2)n^2 + (7/6)n
