Is $\left (\sum_{i=1}^{n}a_{i}b_{i} \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$ Is it true that
$$\left (\sum_{i=1}^{n}a_{i}b_{i}  \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$$
That is, can we always combine sums in that way?
 A: This is true. You can, for example by induction on $m$, prove that
$$\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^m b_j\right)=\sum_{i=1}^n\sum_{j=1}^m a_ib_j,$$
which implies your claim.
If $m=0$, both sides are trivially $0$, so we're good.
Assume that the formula holds for $m$. Let's prove it for $m+1$.
We have
$$\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^{m+1} b_j\right)=\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^{m} b_j+b_{m+1}\right)=\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^m b_j\right)+\sum_{i=1}^na_ib_{m+1}.$$
Now, we apply the induction hypothesis, to obtain
$$\sum_{i=1}^n\sum_{j=1}^m a_ib_j+\sum_{i=1}^na_ib_{m+1}=\sum_{i=1}^n\left(\sum_{j=1}^m a_ib_j+a_ib_{m+1}\right)=\sum_{i=1}^n\sum_{j=1}^{m+1}a_ib_j.$$
A: Direct proof: $$\begin{align}\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j} &\overset 1=\sum_{j=1}^{n}a_{1}b_{1}c_{j}d_{j} +  \sum_{j=1}^{n}a_{2}b_{2}c_{j}d_{j}+\dots +\sum_{j=1}^{n}a_{n}b_{n}c_{j}d_{j} \\&\overset2=a_{1}b_{1}\sum_{j=1}^{n}c_{j}d_{j}+a_{2}b_{2}\sum_{j=1}^{n}c_{j}d_{j} +\dots + a_{n}b_{n}\sum_{j=1}^{n}c_{j}d_{j} \\ &\overset3= (a_1b_1 + a_2b_2 + \dots +a_nb_n)\sum_{j=1}^{n}c_{j}d_{j} \\ &\overset4= \left 
(\sum_{i=1}^{n}a_{i}b_{i}  \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right )\end{align}$$
Explanation:

*

*Expand the first sum (by writing out all terms individually)

*Exclude the factors that are not dependent on index $j$

*Factor out the common factor (the $j$-sum)

*Rewrite the first factor with the sigma notation

A: More generally, $\sum_ix_i\sum_jy_j=\sum_{ij}x_iy_j$, by distributivity in both directions. (However, as @Zuy noted, distributing over a sum still needs induction on the number of terms.) That $x_i=a_ib_i, \, y_j=c_jd_j$ is a distraction.
A: Yes, it is true.
$$\sum_i\sum_j a_ib_j=\sum_i a_i\sum_j b_j=\left(\sum_jb_j\right)\sum_ia_i=\left(\sum_i a_i\right)\left(\sum_j b_j\right)$$
You first take $a_i$ out as common factor which is independent of $j$, then take $\sum b_j$ out as common factor which is independent of $i$.
