How does $\sum_{k=j}^{i+j}(i+j-k)$ = $\sum_{k=1}^{i}(k)$ I am working with summations and I came across these two equivalent summations
$\sum_{k=j}^{i+j}(i+j-k)$ = $\sum_{k=1}^{i}(k)$ but there is no explanation as to how the latter summation was computed from the former.

Express the sum as:
$$\sum_{k=j}^{i+j}(i+j-k)=\sum_{k=j}^{i+j}(i-(k-j))$$
Change the bound:
$$\begin{align}k&=j,j+1,j+2,...,j+i-1,j+i \Rightarrow \\
k-j&=0,1,2,...,i \end{align}$$
Hence:
$$\sum_{k=j}^{i+j}(i-(k-j))=\sum_{k-j=0}^{i}(i-(k-j))=\sum_{t=0}^{i}(i-t)$$
Change the bound:
$$\begin{align}t&=0,1,...,i \Rightarrow \\
-t&=0,-1,...,-i \Rightarrow\\
i-t&=i,i-1,...,0\\
i-t&=0,1,2,...,i-1,i
\end{align}
$$
Hence:
$$\sum_{t=0}^{i}(i-t)=\sum_{i-t=0}^{i}(i-t)=\sum_{k=0}^{i}k=\sum_{k=1}^{i}k.$$
