Simplification of Summation for an Expected Value Problem

Question

Could Someone explain how the expression is simplified here (from 1 to 2)?

$g(x,y) = 3(1+x)y$

(1)$$E[g(X,Y)] = \sum_{k=0}^1 \sum_{l=0}^3 g(k,l)p_{X,Y}(k,l) = \sum_{k=0}^1 \sum_{l=0}^3 3(1 +k)lp_{X,Y}(k,l) $$ (2)$$=\sum_{l=0}^3 3lp_{X,Y}(0,l) + \sum_{l=0}^3 6lp_{X,Y}(0,l)$$

edit: getting rid of the external link (sorry!!)

Answer 1

Since $k=0$ and $k=1$, the sum becomes

$$\sum_{k=0}^1\sum_{\ell = 0}^3 3(1+k)\ell p_{X,Y}(k,\ell) = \sum_{\ell = 0}^3 \left[3 \ell p_{X,Y}(0,\ell) + 3(1+1)\ell p_{X,Y}(1,\ell)\right]$$ $$=$$ $$\sum_{\ell = 0}^3 3\ell p_{X,Y}(0,\ell) + \sum_{\ell = 0}^36\ell p_{X,Y}(1,\ell)$$

simply by using the fact that :

$$\sum(a+b) = \sum a + \sum b$$