Find 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.
+128474 We have these summations.....using the sum of differences
3 11 26 50 85
8 15 24 35
7 9 11
2 2
0
We have 3 rows of non-zero differences until we get to a zero difference row
This will give us a 3rd degree polynomial ax^3 + bx^2 + cx + d
We have the following system
a + b + c + 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
Simplifying this, we get
a + b + c + d = 3
8a + 4b + 2c + d = 11
27a + 9b + 3c + d = 26
64a + 16b + 4c + d = 50
This system isn't difficult to solve, just tedious......I used WolframAlpha to get that
a= 1/3 b = 3/2 c = 7/6 d = 0
So our polynomial is
f(n) = (1/3)n^3 + (3/2)n^2 + (7/6)
