Q:

Double Summation $\sum_{i=1}^{n+1}\sum_{j=0}^{n-1}(i+j)=(2n+1)\sum_{r=1}^n r=3\sum_{r=1}^n r^2$

It can be easily shown by step-by-step and rather messy summation over $j$ and then over $i$ that $$\sum_{i=1}^{n+1}\sum_{j=0}^{n-1}(i+j)=\tfrac12 (2n+1)n(n+1)$$
Note that RHS is equivalent to
$$\displaystyle (2n+1)\sum_{r=1}^n r$$

(1) Is there a clever transform that will simplify the original summation into the summation above without first working out the closed form?

It is interesting to note that RHS is also equivalent to 
$$3\sum_{r=1}^n r^2$$

(2) Is there also another clever transform to convert the original summation into this summation without first working out the closed form? 

A:

(1)
$$\sum_{i=1}^{n+1}\sum_{j=0}^{n-1}(i+j)=\sum_{j=0}^{n-1}\sum_{i=1}^{n+1}i + \sum_{i=1}^{n+1}\sum_{j=0}^{n-1}j=n\left((n+1)+\sum_{i=1}^n i\right)+(n+1)\left(-n+\sum_{j=1}^n j\right) \\ =(n+(n+1))\sum_{r=1}^n r=(2n+1)\sum_{r=1}^n r$$
(2)
$$\sum_{i=1}^{n+1}\sum_{j=0}^{n-1}(i+j)=
\sum_{\begin{array} c 1 \le i \le n+1,\\ 0 \le j \le n-1, \\ 1 \le i+j \le n \end{array}} \!(i+j) + \!\!\sum_{\begin{array} c 1 \le i \le n+1,\\ 0 \le j \le n-1, \\ n+1 \le i+j \le 2n \end{array}} \!((i-1)+(j+1)) \\
=\sum_{k=1}^n\sum_{\begin{array} c 1 \le i \le n+1,\\ 0 \le j \le n-1, \\ i+j=k \end{array}} \!k + \!\sum_{\begin{array} c k=0 \\ (i-1=k) \end{array}}^n\!\!\sum_{\begin{array} c 0 \le j \le n-1, \\ n-k \le j \le 2n-1-k \end{array}} \!\!\!\!k + \!\sum_{\begin{array} c k=1 \\ (j+1=k) \end{array}}^n\!\sum_{\begin{array} c 1 \le i \le n+1, \\ n+2-k \le i \le 2n+1-k \end{array}} \!\!\!\!\!k \\
=\sum_{k=1}^n \,k \!\sum_{\begin{array} c 1 \le i \le k \\ (j=k-i) \end{array}} \!\!1 + \!\sum_{k=1}^n \,k \sum_{j=n-k}^{n-1} \!1 + \!\sum_{k=1}^n \,k\sum_{i=n+2-k}^{n+1} \!1 \\
= \sum_{k=1}^n k \cdot k + \sum_{k=1}^n k \cdot k + \sum_{k=1}^n k \cdot k = 3 \sum_{r=1}^n r^2$$

