Triple sigma($\sum\sum\sum$) question I was solving "multiple sigma" questions, while I stumbled upon this:
Find
$$\sum_{i=0}^n\sum_{j=0}^n\sum_{k=0}^n \frac{1}{3^i3^j3^k} \ ,i\ne j \ne k$$

I'm used to solve double sigma problems (like $\sum_{1\le i<j\le n}\sum i.j$) using triangles like these:

However I'm not sure about how I'm going to construct a cube and splitting it into tetrahedrons(if at all possible)!! Else how do I approach this?
P.S.: The variables are dependent
The answer is $\boxed{\frac{81}{208}}$ when $n\to\infty$
 A: The requitred sum is
$$S=(\sum_{0k=0}^{\infty} 3^{-k})^3-3 \sum_{i=0}^{\infty} 9^{-i} \sum_{j=0} 3^{-j} +2 \sum_{k=0}^{\infty} 3^{-3k}$$
$$\implies S=\left(\frac{1}{1-1/3}\right)^3-3\frac{1}{1-1/9}.\frac{1}{1-1/3}+2\frac{1}{1-1/27}=\frac{27}{8}-\frac{81}{16}+\frac{27}{13}=\frac{81}{208}$$
You may see my answer in this post for the explanation:
$\sum_i \sum_j \sum_k (1 \le i \le j \le k \le n) ~~f_i f_j f_k $ in terms of $\sum f_i~, \sum f^2 _i ~\mbox{and}~ \sum f^3 _i $
A: 
I'm going to construct a cube and splitting it into tetrahedrons(if at all possible)!!

It is not necessary to construct a cube. Let $A = \{0,1,2,...,n\}$. The required combinations (which you mentioned in the square) is basically Cartesian product $A×A$. Extending this further $A×A×A = (A×A)×A$. So write down elements of $A×A$ (which you found by your square) in row and elements of $A$ in column. Combine elements in the rectangle to arrive at the required combinations.
EDIT 2.0 (Approach towards solution)
While the given problem can also be solved using Principle of Inclusion and Exclusion, I am going to solve it using the 'square'/'rectangle' method that I have mentioned earlier. Write $\{ (0,0),(0,1),...,(0,n),(1,0),(1,1),...,(1,n),...,(n,0),(n,1),...,(n,n) \}$ in the horizontal direction and $\{ 0,1,2,...,x,...,n \}$ in the vertical direction. This rectangle can be split into squares formed by Cartesian Product of $\{ (x,0),(x,1),...,(x,n) \}$ and $\{0,1,...,n\}$ for $0\leq x \leq n$. This is shown below:

for $i \neq j \neq k$, we have to find for total $-$(minus) yellow portion. Total $=3^{-x} (\sum_{i=o}^n{3^{-i}})^2$ [solve row-wise and add up all the row-wise results to arrive at this result]. Yellow portion = diagonal + (x,x) column + x row - 2(x,x,x)[since I have deducted (x,x,x) 3 times but want it to be deducted only once]. Diagonal $= 3^{-x} \sum_{i=0}^n{3^{-2i}}$. (x,x) Column $= 3^{-2x} \sum_{i=0}^n{3^{-i}}$. x row $= 3^{-2x} \sum_{i=0}^n{3^{-i}} $ .So, Yellow portion $= 3^{-x} \sum_{i=0}^n{3^{-2i}} + 2 \times 3^{-2x} \sum_{i=0}^n{3^{-i}} -2 \times 3^{-3x}$. Therefore required answer is $$\sum_{x=0}^n{(3^{-x} (\sum_{i=o}^n{3^{-i}})^2-3^{-x} \sum_{i=0}^n{3^{-2i}} -2 \times 3^{-2x} \sum_{i=0}^n{3^{-i}} + 2 \times 3^{-3x})}$$ $$= \frac{27}{8}-\frac{27}{16}-2\times \frac{27}{16} + 2\times \frac{27}{26} = \boxed{\frac{81}{208}}$$
Note: We can solve for $i<j<k$ by reducing the original square into a smaller version $\{ (x,x+1),...,(x,n) \} \times \{x+1,...,n\}$ (see the diagram). The logic here is area of green portion = (area of reduced square -(minus) area of diagonal of reduced square)/2
