Particular solution of recurrence relation

Question

I've got this recurrence relation:

$$M_n = M_{n-1} + n(2n-1)|M_0 = 0$$

and can't think of any form of particular solution to get a solvable constant.

With $M_n^H = K$being the homogeneous part of the equation (and $K$ being a constant), obviously, I can't seem to undertand what I should try.

I've tried a few particular solutions ($M_n^P$) to no avail. Am I missing something? Here is what I chose (A and B being the constant(s) to determine)

1. $$M_n^P = An+Bn^2$$ $$An+Bn^2 = A(n-1) + B(n-1)^2 + n(2n-1)$$ $$An+Bn^2 = An - A + Bn^2-2Bn+B+2n^2-n$$ $$0 = (B-A) + (-2B-1)n + (2)n^2$$ $$0+ 0n + 0n^2 = (B-A) + (-2B-1)n + (2)n^2$$

This means $B\neq A$ and $B = \frac{-1}{2}$. ...And this is where I get really confused and feel like I hit a wall. Am I in the right direction? Or is assuming $(-2B-1) = 0$ false? Am I just missing the obvious?

The complete solution should be : $$M_n = \frac{n(n+1)(4n-1)}{6}$$

Answer 1

1. As well as the $An^2+Bn$ you need a lower order term: $An^2+Bn+C$. But. . .
2. since $C$ is a solution of the homogeneous case, you actually need $An^3+Bn^2+Cn$.

Enjoy the algebra ;-)