1
$\begingroup$

Let $a_{i,j}$ any positive integers wher $1\leq i,j \leq 3$.

I want to write closed form for following summation:

$$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$

$\textbf{My attempt:}$ \begin{equation*} \sum_{j=1}^3 a_{j,3}\left[\prod_{\substack{i=1 \\ i\neq j}}^3\left(\sum_{\substack{k=1 \\ k<j}}^{3 \ \ \ \text{or} \ \ 3-1} a_{i,k} \right) \right] \end{equation*} We will take $3-1$ when $k<j$.


Actually, I want to generalize this. My attempt so complicated to me.

If we consider all of $a_{i,j}$ as a matrix: \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} then are there any computer program to make formula?


$$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})\cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+\cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+\cdots (a_{r,1}+a_{r,2}+a_{r,3}) \\ \vdots \\ +a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})\cdots (a_{r-1,1}+a_{r-1,2}) $$

$\endgroup$
3
$\begingroup$

$$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$ $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$ $$=\prod_{i=1}^3\sum_{j=1}^3a_{i,j}-\prod_{i=1}^3\sum_{j=1}^2a_{i,j}.$$ More generally, $$\sum_{i=1}^ra_{i,m}\left(\prod_{h=1}^{i-1}\sum_{j=1}^{m-1}a_{h,j}\right)\left(\prod_{h=i+1}^r\sum_{j=1}^ma_{h,j}\right)=\prod_{i=1}^r\sum_{j=1}^ma_{i,j}-\prod_{i=1}^r\sum_{j=1}^{m-1}a_{i,j}$$ $$=\sum_{m\in\{j_1,j_2,\dots,j_r\}}a_{1,j_1}a_{2,j_2}\cdots a_{r,j_r}.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the answer. Is it possible to generalize it? I edited my question. $\endgroup$ – 1Spectre1 Nov 27 '18 at 13:24
  • 1
    $\begingroup$ @1ENİGMA1 I cleaned up my answer. $\endgroup$ – bof Nov 28 '18 at 23:13
  • 1
    $\begingroup$ For example, $n!=\prod_{i=1}^ni$ and in particular $0!=\prod_{i=1}^0i=1$. Likewise, $x^n=\prod_{i=1}^nx$, in particular, $x^0=\prod_{i=1}^0x=1$. $\endgroup$ – bof Nov 29 '18 at 8:59
  • 1
    $\begingroup$ $\sum_{m\in\{j_1,j_2,\dots,j_r\}}a_{1,j_1}a_{2,j_2}\cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}\cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,\dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set $\{j_1,j_2,\dots,j_r\}$. $\endgroup$ – bof Nov 29 '18 at 9:04
  • 1
    $\begingroup$ $$\prod_{i=1}^2\sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$\sum_{2\in\{j_1,j_2\}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2\notin\{j_1,j_2\}=\{1,1\}$. $\endgroup$ – bof Nov 30 '18 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.