How to sum like this

Question

I want to sum as follows, where j is from 1 to n in each sum (if s is added to 2 when j is not added to n, then j does not need to be n):

$$Table[\sum_{s=k,j=1}^{n}(s+j),\lbrace{k,0,n\rbrace}]$$ We should get the following results (if n = 2):

{9,6,3}

I can achieve this goal like this, but is there any better way?

Table[(j = 0; Sum[(j = j + 1; (s + j)), {s, k, 2}]), {k, 0, 2}]

In fact, I have to use the sum symbol sum to do something similar to the following figure:

I deal with it like this, but I wonder if there is any other way:

boundaryD[kernelF_, x_, f_, n_: 1] :=

 Module[{j}, 
  Table[(j = 0; 
    Sum[(j = j + 1; (-1)^(j - 1)
        D[D[kernelF, D[f[x], {x, s}]], {x, j - 1}]), {s, k, n}]), {k, 
    1, n}]]
boundaryD[F[x, y[x], y'[x], y''[x]], x, y, 2]

Answer 1

ClearAll[tbl]
tbl[n_] := Array[(n + 1) (n + 2 - #) &, n + 1]

tbl[2]

{9, 6, 3}

tbl[5]

{36, 30, 24, 18, 12, 6}

Alternatively,

tbl2[n_] := Table[(n + 1) (n + 2 - i), {i, n + 1}]

Answer 2

An alternative way of getting the same output is this:

Range[(n + 1)^2, n + 1, -n - 1]

or equivalently:

Table[i, {i, (n + 1)^2, n + 1, -n - 1}]

not sure it's what you are looking for though!