Is $\sum_ix_iy_i=\sum_ix_i\sum_iy_i$? I ask this because the equation for the center of mass of a system (made up of a number of small masses attached to each other) is given by:
$$\bar x=\frac{\sum_im_ix_i}{\sum_im_i}$$
If the operation in the question is valid then $\sum_ix_i$ would cancel and just leave $\sum_im_i$ which must be wrong. However, I don't know why the operation in the question is wrong. Can you explain?
 A: $(x_1+x_2)(y_1+y_2) = x_1y_1+x_2y_2+x_1y_2+x_2y_1$: you have the "cross terms" $x_iy_j$ for $i\neq j$. This generalizes easily to $n$ instead of 2. 
A: Certainly not. This would imply, for instance that $x_1^2+x_2^2=(x_1+x_2)^2$.
A: Suppose every $x_i = c$. Then
$$\sum x_i y_i = \sum cy_i = c\sum y_i \neq \sum c \sum y_i.$$
A: No, they are not the same.  Let $x_1=\ldots=x_n=1$ and $y_=\ldots=y_n=1$.
Then 
$$\sum_{i=1}^nx_iy_1=\sum_{i=1}^n1=n$$
but
$$\sum_{i=1}^nx_i\sum_{i=1}^ny_i=\sum_{i=1}^nx_i\sum_{i=1}^n1=\sum_{i=1}^nx_in=n\cdot n=n^2.$$
A: Clearly not, you can examine the correctness of the summation by putting some numbers in it.
$$\begin{array}{l}\sum\limits_i {{x_i}} {y_i}\left( {\begin{array}{*{20}{c}} = \\ \ne \end{array}} \right)\sum\limits_i {{x_i}} \sum\limits_i {{y_i}} \\{x_1} = 1,{x_2} = 2,{x_3} = 3;\\{y_1} = 4,{y_2} = 5,{y_3} = 6;\\\sum\limits_i {{x_i}} {y_i} = {x_1}{y_1} + {x_2}{y_2} + {x_3}{y_3} = 1 \times 4 + 2 \times 5 + 3 \times 6 = 32\\\left. {\begin{array}{*{20}{c}}{\sum\limits_i {{x_i}}  = {x_1} + {x_2} + {x_3} = 1 + 2 + 3 = 6}\\{\sum\limits_i {{y_i}}  = {y_1} + {y_2} + {y_3} = 4 + 5 + 6 = 15}\end{array}} \right| \Rightarrow \sum\limits_i {{x_i}} \sum\limits_i {{y_i}}  = 90\\\left( {\sum\limits_i {{x_i}} {y_i} = 32} \right) \ne \left( {\sum\limits_i {{x_i}} \sum\limits_i {{y_i}}  = 90} \right)\end{array}$$
