Why is $\sum_{k=j}^{i+j}(j+i-k) = \sum_{k=1}^{i}(k)$

Question

$\displaystyle\sum_{k=j}^{i+j}(j+i-k) = \displaystyle\sum_{k=1}^{i}(k)$

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.

Answer 1

\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}

Answer 2

Write out the terms in the first sum: $$\sum_{k=j}^{i+j} (j+i-k)=i+(i-1)+\cdots + 1 + 0 = \sum_{k=1}^i k$$