Equating summations of an integer partition and its conjugate

Question

I've been given an integer partition A = (A₁, A₂, ... ,A_n) and its conjugate B = (B₁, B₂, ... ,B_m). Using that information, I'm tasked with proving that

$$\sum_{i=0}^n (i-1)A_i = \sum_{j=1}^m B_k(B_k -1)(1/2)$$

From observing the pattern of the summations, I've observed a few facts:

1) n = B₁

2) m = A₁

3) the first element of A will not be included in the summation. This is because when i=1, we'll have (0*A₁).

And this is all I've been able to figure it out in the last hour of staring and pondering the problem. If anyone has a suggestion on how to proceed or a resource to look at to further my solution, I'd deeply appreciate it.

Answer 1

If you're a visual thinker then drawing some Ferrers diagrams and marking rectangles which correspond to $(i-1)A_i$ and triangles which correspond to $\frac{B_j(B_j-1)}{2}$ will probably help you see a way forward.

If you prefer to bash away at algebra, you could try taking $B_j = |\{i : A_i \ge j\}|$ and see how you can work that into the sums.