I see this way and idea from Simply Beautiful Art
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Let : $S=\displaystyle\sum_{1≤k≤m≤n≤\infty}f(m)f(k)f(n)$
then :
$6S=\displaystyle\sum_{n,m,k≥1}f(m)f(k)f(n)+3\displaystyle\sum_{n,m≥1}f(m)f^{2} (n)+2\sum_{n≥1}f^{3}(n)$
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now let : $A=\displaystyle\sum_{1≤i≤≤k≤j≤m≤\infty}f(m)f(k)f(j)f(i)$
then $24A=?$
I need understand this brilliant way
There are $4!=24$ permutations of $f(m)f(k)f(j)f(i)$, so we will always need to end up with 24 combinations by adding the following sums:
\begin{align}S_1&=\sum_{m,k,j,i\ge1}f(m)f(k)f(j)f(i)\\S_2&=\sum_{k,j,i\ge1}f(k)f(j)f(i)^2\\S_3&=\sum_{j,i\ge1}f(j)^2f(i)^2\\S_4&=\sum_{j,i\ge1}f(j)f(i)^3\\S_5&=\sum_{i\ge1}f(i)^4\end{align}
Consider the case of all distinct values i.e. $m<k<j<i$. All 24 combinations are then contained within $S_1$.
Consider the case of 3 distinct values i.e. $m=i$ and $j,k$ are distinct. For example, consider $f(1)f(2)f(3)^2$. This appears 12 times in $S_1$ as
$$(m,k,j,i)\in\left.\begin{cases}(1,2,3,3),(1,3,2,3),(1,3,3,2),(2,1,3,3),(2,3,1,3),(2,3,3,1),\\(3,1,2,3),(3,1,3,2),(3,2,1,3),(3,2,3,1),(3,3,1,2),(3,3,2,1))\end{cases}\right\}$$
and in $S_2$, they appear 2 times as
$$(k,j,i)\in\{(1,2,3),(2,1,3)\}$$
and since this leaves us with 12 combinations to reach 24, we need 6 copies of $S_2$.
Consider the case of 2 distinct values with 2 variables for each value i.e. $m=i\ne k=j$. For example, consider $f(1)^2f(2)^2$. There are 6 ways for this to appear in $S_1$:
$$(m,k,j,i)\in\{(1,1,2,2),(1,2,1,2),(1,2,2,1),(2,1,1,2),(2,1,2,1),(2,2,1,1)\}$$
and 2 ways for this to appear in $S_2$:
$$(k,j,i)\in\{(1,1,2),(2,2,1)\}$$
and 2 ways for this to appear in $S_3$:
$$(j,i)\in\{(1,2),(2,1)\}$$
and since we have 6 copies of $S_2$, this leaves us with 6 remaining combinations to reach 24, hence we have 3 copies of $S_3$.
Consider the case of 2 distinct values with 1 and 3 variables for each i.e. $j\ne m=k=i$. For example, consider $f(1)f(2)^3$. This appears 4 ways in $S_1$:
$$(m,k,j,i)\in\{(1,2,2,2),(2,1,2,2),(2,2,1,2),(2,2,2,1)\}$$
and 2 ways in $S_2$:
$$(k,j,i)\in\{(1,2,2),(2,1,2)\}$$
and no ways in $S_3$ and only 1 way in $S_4$. This leaves us with 8 more, hence we have 8 copies of $S_4$.
Consider the case of all equal variables i.e. $m=k=j=i$. For example, consider $f(1)^4$. This appears exactly once in every sum, which thus leaves us with 12 copies of $S_5$.
Hence we have