Summation transformation formula(s) According to this link
$$
\sum\limits_{k=j}^{i+j} (j+i-k)
$$
is
$$
\sum\limits_{k=1}^i (k)
$$
Can someone write the Sum formula(s) which was used in this transformation?
 A: Writing it, 
\begin{align*}
 &\;\;\; \sum\limits_{k=j}^{i+j} (j+i-k) \\ 
 &= ((j + i - j) + (j + i - j - 1) + \dots + 1 + 0)\\
 &= (i + (i-1) + ... +  1) \\
 &= \sum_{k=1}^i k
\end{align*}
Shifting index $k \to k + j$,
$$\sum\limits_{k=j}^{i+j} (j+i-k)= \sum_{k=0}^i (j + i - (k+j)) = \sum_{k=0}^i (i-k) = \sum_{k=1}^i k$$ 
A: Let's do things a bit differently; maybe that will make it clearer.
All that happened here is re-indexing of the terms of the sequence.  In other words, all of the same terms appear... we're just assigning them a different index number, and possibly re-ordering them.
You started off here with
$$
\sum_{k=j}^{i+j}(j+i-k).
$$
Notice that $j+i$ never changes.  Let's define a new index of summation, say $t$, by $t=j+i-k$.
Notice that when we plug in $k=j$, we get $t=i$; when we plug in $k=j+1$, we get $t=i-1$; and so on, and so forth, until we get to $k=j+i$, in which case $t=0$.
So, summing over all values of $k$ between $j$ and $j+i$ is the same as summing over all values of $t$ between $0$ and $i$. All that remains is to rewrite our index $k$ term of the summand in terms of $t$; since $t=j+i-k$, this term is precisely $t$. So, we conclude that
$$
\sum_{k=j}^{j+i}(j+i-k)=\sum_{t=0}^{i}t.
$$
Does that help?
