Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want evaluate the sum. I'm not sure if I'm doing it the right way.

$\sum\limits_{1\le i<j\le 3}{ij} $

my answer : $1(2)+2(3)$

$i$ starts at $1$, $j$ starts at $2$, I multiply them, then $i$ goes to $2$ and $j$ to $3$ and we stop.



share|cite|improve this question
Why not (1)(3)? – Inquest Jan 29 '13 at 0:26
up vote -1 down vote accepted

You need to calculate all the possibles values of $ij$ satisfying the condition $1\le i < j \le 3$. Think of it as a loop -

$i$ starts from $1$ and $j$ from $2$.

$\implies S = (1)(2)$

Now we increment $j$ by $1$.

$\implies S = (1)(2)+ (1)(3)$

According to the condition $1\le i < j \le 3$, $j$ cannot be increment anymore. So we increment $i$ by $1$. $\because i<j, $ $ \therefore j$ now becomes $3$.

$\implies S= (1)(2) + (1)(3) + (2)(3)$

Now if we increment $i$ by one $j$ must be $4$ or greater. This is against the condition, $1\le i < j \le 3$. Thus, neither $i$ nor $j$ can be incremented. This means we have reached end of the loop.

$\therefore S = (1)(2)+(1)(3)+(2)(3) = 11$

share|cite|improve this answer
Well done to break a nut you use tractor – Milingona Ana Jan 29 '13 at 19:27

$$\sum\limits_{1\le i<j\le 3}{ij}=\sum_{i=1}^{2}\left(\sum_{j=i+1}^{3}ij\right)=\sum_{j=1+1}^{3}1j+\sum_{j=2+1}^{3}2j= $$ $$=\sum_{j=2}^{3}1j+\sum_{j=3}^{3}2j=1\cdot2+1\cdot3+2\cdot3=11 $$

share|cite|improve this answer

No, you have not understood the notation. It just means adding up all possible evaluations of ij under the restriction that $1\le i<j\le 3$.

share|cite|improve this answer

You need $1\times 2+1\times 3+2\times 3$.

share|cite|improve this answer

It should be $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3$.

share|cite|improve this answer

$1 \cdot 2+1 \cdot3+2\cdot3=11$

share|cite|improve this answer

Although multiple people have shown how to answer this question by enumerating the terms, since this is labelled homework I assume you would also like to know the general technique.

If you think of the two summations as summing across the rows and columns of a matrix, here labelled with the $i,j$ values: $$\begin{matrix} 11 & 12 & 13 \\ 21 & 22 & 23 \\ 31 & 32 & 33 \\ \end{matrix} $$ you will see that the terms you are asked to sum, given $i \lt j$, are in the upper right triangle. For any function of $i$ and $j$ where $i$ and $j$ are interchangable, the sum over this triangle will equal the sum over the lower left triangle. Now you can see that the total of the whole matrix is $$M = UT + LT +D = 2 \cdot UT + D$$ (M=matrix, UT=upper triangle, D=diagonal), therefore the sum over the upper triangle is $$ \frac{M - D}2$$

Summing over the whole matrix requires a lot less bookkeeping, and in your case reduces to $$M = \sum_{i=1}^3 i \cdot \sum_{j=1}^3 j = 6 \cdot 6 = 36$$ $$D = \sum_{i=1}^3 i^2 = 14$$ (note that this came from setting $j = i$ thus $i \cdot j = i^2$) and $$\frac{M-D}2 = \frac{36-14}2 = \frac{22}2 = 11$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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