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What is the value of $\sum \limits_{1 \le i < j \le 10} ij$?

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*hint*$$\left ( \sum _{k=1}^{n}k \right )^2=\sum _{k=1}^{n}k^2+2 \sum_{1 \leq i < j \leq n}ij$$

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How did you deduce this expression? – Ambesh Apr 16 '13 at 21:28

$$\sum_{1\le r\le n}r\left(\sum_{r+1\le s\le n}s\right)$$ $$=\sum_{1\le r\le n}r\{\sum_{1\le s\le n}s-\sum_{1\le s\le r}s\}$$ $$=\sum_{1\le r\le n}r\{\frac{n(n+1)}2-\frac{r(r+1)}2\}$$

$$=\frac{n(n+1)}2\sum_{1\le r\le n}r- \frac12\{\sum_{1\le r\le n}r^3+\sum_{1\le r\le n}r^2\}$$

Now, you know $\sum_{1\le r\le n}r=\frac{n(n+1)}2,$

$\sum_{1\le r\le n}r^2=\frac{n(n+1)(2n+1)}6,$

$\sum_{1\le r\le n}r^3=\left(\frac{n(n+1)}2\right)^2$

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$\sum \limits_{1 \le i < j \le 10} ij = \frac 1 2 \left({\left(\sum \limits_{1 \le i \le 10} i\right)^2} - \sum \limits_{1 \le i \le 10} i^2 \right)= \frac 1 2 (55^2 - \frac{10 \cdot 11 \cdot 21}{6}) = 1320$

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