Separated 3Sum versus 3Sum problem Does it matter in the 3Sum problem if the numbers to be summed belong to the same set or to distinct sets?

Let's define 


the problem "$k$-Sum" as follows: given a single finite set of integers $X\subset N$ of size $n$, decide if there exists $k$ integers $x_1,\ldots,x_k\in X$ such that their sum $\sum_{i=1}^k x_i$ is null; and 
the problem "Separated $k$-Sum" as follows: given $k$ finite sets of integers $X_1,\ldots,X_k\subset N$ of respective sizes $n_1,\ldots,n_k\in[1..n]\subset N$, decide if there exist a vector $(x_1,\ldots,x_k)\in X_1\times\ldots\times X_k$ such that their sum $\sum_{i=1}^k x_i$ is null.  


I was assuming that those two problems were equivalent, and I was using the second one as a pedagogical example for some parameterization technique (while there is an algorithm running in $O(n^2)$ for both, the second one can take advantage of variations in the size of the sets to solve the problem much faster, down to $O(n_1 n_3+\sum n_i\lg n_i)$ and $O(n_1 n_2 \lg \frac{n_3}{n_2}+\sum n_i\lg n_i)$).

But I realized today that I did not know how to prove this equivalence, and I might even start to think that they are not.
You can reduce the SEPARATED k-SUM problem to a (k+1)-SUM problem in the following way:

Given $X_1,...,X_n$, let $b$ such that $2^b &gt; \max( abs(X_1) \cup ... \cup \;abs(X_n))$, where $abs(X_i) = \{ |x| \mid x \in X_i \}$, and let $c$ such that $2^c &gt; k 2^b$ 

For  $i = 1,...,k$ build $Y_i = \{ 2^{c+ki} + 2^{b} + x_j \mid x_j \in X_i \}$;
let $u = - \sum_{i=1}^k 2^{c+ki} -k2^b$

Build an equivalent (k+1)-SUM problem picking $X =  \{ u \} \cup Y_1 \cup ... \cup Y_k$

Note that every $Y_i$ contains only positive elements.

Informally we require a $k+1$ elements subset $C$ from $X$ whose elements sum to zero.
We cannot build a solution only with the negative element $u$; so we must include in $C$ at least one element $y$ from one of the $Y_i$; but the $2^{c+ki}$ "component" of $y$ can be balanced only if $u$ is also included; but including $u$ implies that exactly one of the elements of each $Y_i$ is included (the  $- \sum_{i=1}^k 2^{c+ki}$ component of $u$ allows to keep the separation between the elements of the original $X_i$). But we also have $\sum_{j=1}^k (2^b + x_j) = k2^b$ and this implies that the original $x_j$s (coming from distinct $X_i$s as seen above) sum to zero.