+1710 The sum
6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right)
is equal to a polynomial f(n) for all n \ge 1.
Write f(n) as a polynomial with terms in descending order of n.
+129839 \(6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) \)
For the moment, disregard the "6"
We have
1 + 4 + 9 + 16 + 25 + 36
1 5 14 30 55 91
4 9 16 25 35
5 7 9 11
2 2 2
We have three rows of non-zero differences
The generating polynomial willl be
an^3 + bn^2 + cn + d
We have this system
a + b + c + d = 1
8a + 4b + 2c + d = 5
27a + 9b + 3c + d = 14
64a + 16b + 4 + d = 30
Solving this produces
a = 1/3 b = 1/2 c =1/6 d = 0
The polynomial is
(1/3)n^3 + (1/2)n^2 + (1/6)n
Multiplying by 6 we get
f(n) = 2n^2 + 3n + n
