# Summation of $i \cdot j$ from $1$ to$3$

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.

Right?

Thanks.

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Why not (1)(3)? – Inquest Jan 29 '13 at 0:26

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.

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

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Well done to break a nut you use tractor – Milingona Ana Jan 29 '13 at 19:27

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 $$\frac{M-D}2 = \frac{36-14}2 = \frac{22}2 = 11$$

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

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

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

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

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

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