Summation Manipulation Problem Is the following statement true:
$$\sum^k_{j=0} \left(\sum^j_{i=0}a_i b_{j-i}\right) d_{k-j} = \sum^k_{j=0} \left(\sum^j_{i=0}b_i d_{j-i}\right) a_{k-j}$$
I'm not sure if this is true as I've been unable to prove this directly. However, I think this is true. 
If we fix $k$ and then $j$, we have fixed the subscript of $d$. So, there is a term $a_m b_n d_o$ on the LHS such that $m + n + o = k$.
However on the RHS, if we fix the same value of $k$ as above and strategically fix a different value of $j$ so that the subscript of $a$ is $m$, then we can find a particular $i$ such that we also have $a_m b_n d_o$. That is, the subscript of $b$ plus the subscript of $d$ on the RHS is equal to $j$, so we can find an $n$ and $o$ such that $n+o = k - m = j$.
If it is true, is there a more direct (or different) proof of this? If not, why is it not true?

Both sides equalize: $$\sum_{m+n+o=k}a_mb_nd_o$$ where $m,n,o$ denote nonnegative integers.
The expression: $$\sum_{m+n+o=k}a_mb_nd_o$$ abbreviates: $$\sum_{\langle m,n,o\rangle\in S}a_mb_nd_o$$ where $S=\{\langle m,n,o\rangle\in\mathbb N\mid m+n+o=k\}$ and $\mathbb N=\{0,1,2,\dots\}$.
Further $|S|=\binom{k+2}2$ (which can be found with stars and bars).
