Show that $\sum\limits_{r=1}^n\sum\limits_{k=1}^{n-r+1}(2r-1)k=\frac 13\sum\limits_{r=1}^nr\sum\limits_{s=1}^{n+1}s$ 
Question:
  Show that 
  $$\sum_{r=1}^n\sum_{k=1}^{n-r+1}(2r-1)k=\frac 13 \sum_{r=1}^nr\sum_{s=1}^{n+1}s$$
  purely by manipulating summation limits and summands, i.e. convert the original summation (on the left) to the summation on the right, without first evaluating the closed form, as shown below.

This question is related to Question $2435490$ on MSE.
From the solutions provided , it is clear that 
$$\sum_{r=1}^n\sum_{k=1}^{n-r+1}(2r-1)k=\frac 1{12}n(n+1)^2(n+2)$$
This can also be written as 
$$\frac 13\cdot  \frac {n(n+1)}2\cdot\frac {(n+1)(n+2)}2$$
which is the closed form of
$$\frac 13 \sum_{r=1}^nr\sum_{s=1}^{n+1}s$$

Not a solution. 
Consider $n=4$
$LHS = 1(1+2+3+4) + 3(1+2+3) + 5(1+2) + 7 = 1(1+3+5+7) + 2(1+3+5) + 3(1+3)+ 4 = \sum\limits_{k=1}^n (n+1 - k)k^2 = \sum\limits_{k=1}^n (n+1)k^2 - k^3 $  
Separately, $$\frac{1}{3}\sum\limits_{k=1}^n (n+1)k + k^3 = \frac{1}{3}(n+1)\sum\limits_{k=1}^n k + \frac{1}{3}(\sum\limits_{k=1}^n k)^2$$
$$= \frac{1}{3}(\sum\limits_{k=1}^n k )(n + 1 + \sum\limits_{k=1}^n k) $$
$$= \frac{1}{3}(\sum\limits_{k=1}^n k )(\sum\limits_{k=1}^{n+1} k)$$
So now all that remains is to prove: 
$$\sum\limits_{k=1}^n ((n+1)k^2 - k^3)  = \frac{1}{3}\sum\limits_{k=1}^n ((n+1)k + k^3)$$
Hope this helps a bit. 
