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I'm practising with precalculus exercises and came across this question of double summation. But I couldn't do it like a single sigma notation. Question is, how can I write this formulae in full to see what it does.

$$\sum_{i=1}^3 \sum_{j=1}^2 w_i x_j$$

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Handle the inner summation first. \begin{align} \sum_{i=1}^3 \sum_{j=1}^2 w_i x_j &= \sum_{i=1}^3 (w_i x_1+w_ix_2)\\ \end{align} Are you able to do it now?

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  • $\begingroup$ not really... even if I can, can't be sure if it is true. $\endgroup$ – PRCube Oct 25 '16 at 2:40
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    $\begingroup$ @PRCube $$= (w_1x_1 + w_1x_2) + (w_2x_1 + w_2x_2) + (w_3x_1 + w_3x_2)$$ $\endgroup$ – user137731 Oct 25 '16 at 3:11
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Here is another variation which might be helpful. We start with a simpler example for demonstration. Using the distributive law once without sigma notation and once with sigma notation we can write \begin{align*} w(x_1+x_2)&=wx_1+wx_2\\ w\sum_{i=1}^2x_i&=\sum_{i=1}^2wx_i \end{align*}

We obtain \begin{align*} \sum_{i=1}^3\sum_{j=1}^2w_ix_j&=\sum_{i=1}^3w_i\sum_{j=1}^2x_j\tag{1}\\ &=\left(\sum_{i=1}^3w_i\right)\left(\sum_{j=1}^2x_j\right)\\ &=(w_1+w_2+w_3)(x_1+x_2) \end{align*}

Since $w_i$ is not depending on the inner sum, we can factor it out in (1).

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