Q:

Prove this sum is always zero

I encountered this problem in the context of singular homology, trying to prove that the boundary map is always zero.
How do you show that the following sum, for all $n\in\mathbb{N}$, is always zero?
$$ \left[\sum_{k=0}^{n-1}\sum_{j=0}^k(-1)^{j+k} A_{k+1,j} \right]+\left[\sum_{k=0}^{n-1}\sum_{j=k+1}^{n}(-1)^{j+k}A_{j,k}\right] $$ where $A_{l,m}$ are some numbers which depend on two indices.
I have written out the first few terms of each sum and verified that it is indeed zero, but I would like to make a clean change of variable that somehow clearly shows this. I would also like to avoid induction on $n$.

A:

$$ \left[\sum_{k=0}^{n-1}\sum_{j=0}^k(-1)^{j+k} A_{k+1,j} \right]= \left[\sum_{k=1}^{n}\sum_{j=0}^{k-1} (-1)^{j+k-1} A_{k,j} \right] $$
Let's change the letters of indice (just a change of name):
$$= \left[\sum_{j=1}^{n}\sum_{k=0}^{j-1} (-1)^{j+k-1} A_{j,k} \right] $$
Let's invert both sums :
$$= \left[\sum_{k=0}^{n-1}\sum_{j=k+1}^{n} (-1)^{j+k-1} A_{j,k} \right]$$
And finally
$$= - \left[\sum_{k=0}^{n-1}\sum_{j=k+1}^{n}(-1)^{j+k}A_{j,k}\right] $$

