Why is $\sum_{k=j}^{i+j}(j+i-k) = \sum_{k=1}^{i}(k)$ $\displaystyle\sum_{k=j}^{i+j}(j+i-k) = \displaystyle\sum_{k=1}^{i}(k)$
I know the above are equal through testing it out with arbitrary values, but I can't get an intuitive grasp as to why this is. 

\begin{align}
S &= \sum_{k=j}^{i+j} (i+j-k) \\
&= (i) + (i-1) + \cdots + ((i+j)-(i+j-1)) + ((i+j)-(i+j)) \\
&= (i) + (i-1) + \cdots + 1 + 0 \\
&= 0 + 1 + \cdots + (i-1) + i \\
&= \sum_{k=0}^{i} k.
\end{align}
Also 
\begin{align}
\sum_{n=1}^{m} n = \frac{m(m+1)}{2}
\end{align}
such that
\begin{align}
\sum_{k=j}^{i+j} (i+j-k) = \sum_{k=1}^{i} k = \binom{i+1}{2}.
\end{align}
