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?

Hint:
With $k=2$,
$$(a_0b_0)d_2+(a_0b_1+a_1b_0)d_1+(a_0b_2+a_1b_1+a_2b_0)d_0$$
vs.
$$(d_0b_0)a_2+(d_0b_1+d_1b_0)a_1+(d_0b_2+d_1b_1+d_2b_0)a_0.$$
All terms are matching. There are indeed $k(k+1)/2$ of them (triangular number), such that the sum of indices is $k$.
