Triple summation having $i,j,k$ as variables

Question

Finding value of

$(1)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$

$(2)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$

For $(1) $ what I have done as

For without any restriction as

$ (S)=\displaystyle \mathop{\sum\sum\sum}_{1\leq i,j,k\leq n}ijk=\bigg(\sum^n_{i=1}i\bigg)^3=\bigg(\frac{n(n+1)}{2}\bigg)^3$

For exactly one equal

$\displaystyle S'=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j,k\leq n}ijk=\bigg(\sum^n_{j=1}j^2\bigg)\bigg(\sum^n_{k=1}k\bigg)=\bigg(\frac{n(n+1)(2n+1)}{6}\bigg)\bigg(\frac{n(n+1)}{2}\bigg)$

There are 3 such pairs $( i=j,j=k,k=i)$

All three equal

$S''=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j=k\leq n}ijk=\sum^n_{k=1}k^3=\bigg(\frac{n(n+1)}{2}\bigg)^2$

Now we have

$\displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk=S-3S'+S''$

$\displaystyle =\frac{n^2(n+1)^2}{4}\bigg(\frac{3n^2-n+4}{6}\bigg)$

But this is not correct, please tell me where I am wrong

Also please tell me part $2$

Answer 1

It is better to make each summation explicit. You can then evaluate them one by one. You are looking for $$\sum_{k=2}^n\sum_{j=2}^{k}\sum_{i=1}^{j-1}ijk$$ where the $2$ comes because we require $i \lt j$. Now evaluate from the inside out, so this becomes $$\sum_{k=2}^n\sum_{j=2}^{k}\sum_{i=1}^{j-1}ijk=\sum_{k=2}^n\sum_{j=2}^{k}\frac{j(j-1)}2jk$$ Keep going. For the second one, you just update the lower limits to $1$ and the upper $i$ limit to $j$.

Answer 2

Your approach is smart. It just needs a little revision. Based on Faulhaber's formula we use \begin{align*} S_n&=\sum_{1\leq i,j,k\leq n}ijk=\left(\sum_{i=1}^ni\right)^3=\frac{1}{8}n^3(n+1)^3\\ S^{\prime}_n&=\sum_{1\leq i=j,k\leq n}ijk=\left(\sum_{j=1}^{n}j^2\right)\left(\sum_{k=1}^nk\right)\\ &=\frac{1}{6}n(n+1)(2n+1)\frac{1}{2}n(n+1)\\ &=\frac{1}{12}n^2(n+1)^2(2n+1)\\ S^{\prime\prime}_n&=\sum_{1\leq i=j=k\leq n}ijk=\sum_{i=1}^ni^3=\frac{1}{4}n^2(n+1)^2\\ \color{blue}{S^{\prime\prime\prime}_n}&\color{blue}{=\sum_{1\leq i<j<k\leq n}ijk}\\ \end{align*}

The identity we use to get $\color{blue}{S^{\prime\prime\prime}_n}$ is \begin{align*} \color{blue}{S_n=6\cdot S^{\prime\prime\prime}_n+3\cdot S^{\prime}_n-2\cdot S^{\prime\prime}_n}\tag{1} \end{align*}

Note, in (1)

• we use $6$ times $S^{\prime\prime\prime}_n$ since we have $3!=6$ permutations of $(i,j,k)$ to respect.

• we also have to consider that $1\leq i=j,k\leq n$ also contains the cases $1\leq i=j\color{blue}{=}k\leq n$, which are also counted when we sum over $1\leq i=k,j\leq n$ and $1\leq i, j=k\leq n$. To compensate this overcounting we subtract $S^{\prime\prime}_n$ twice.

We obtain according to (1) and the identities above \begin{align*} \color{blue}{S^{\prime\prime\prime}_n}&=\frac{1}{6}S-\frac{1}{2}S^{\prime}+\frac{1}{3}S^{\prime\prime}\\ &=\frac{1}{48}n^3(n+1)^3-\frac{1}{24}n^2(n+1)^2(2n+1)+\frac{1}{12}n^2(n+1)^2\\ &=\frac{1}{48}n^2(n+1)^2\left(n(n+1)-2(2n+1)+4\right)\\ &=\frac{1}{48}n^2(n+1)^2\left(n^2-3n+4\right)\\ &\,\,\color{blue}{=\frac{1}{48}n^2(n+1)^2(n-1)(n-2)} \end{align*} in accordance with a comment from @JG.

The other wanted sum (2) can be derived from $S^{\prime\prime\prime}_{n}$. We obtain \begin{align*} \color{blue}{\sum_{1\leq i\leq j\leq k\leq n}}\color{blue}{ijk} &=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j<k\leq n}ijk+\sum_{1\leq i<j=k\leq n}ijk\\ &\qquad\qquad+\sum_{1\leq i=j=k\leq n}ijk\\ &=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j<k\leq n}ijk+\sum_{1\leq k<i=j\leq n}ijk\\ &\qquad\qquad+\sum_{1\leq i=j=k\leq n}ijk\\ &=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j,k\leq n}ijk\\ &\,\,\color{blue}{=S^{\prime\prime\prime}_n+S^{\prime}_n}\\ &=\frac{1}{48}n^2(n+1)^2(n-1)(n-2)+\frac{1}{12}n^2(n+1)^2(2n+1)\\ &=\frac{1}{48}n^2(n+1)^2(n^2+5n+6)\\ &\,\,\color{blue}{=\frac{1}{48}n^2(n+1)^2(n+2)(n+3)} \end{align*} in accordance with a comment from @JG.