Prove that the number of answers for $|a_1|+|a_2|+...+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n| \le k$. I'm trying to solve this problem:

Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|≤k$. all $a_i$ is an integer number.

I have a solution for non negative $a_i$:
We convert $|a_1|+|a_2|+...+|a_k|≤n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|≤k$ to $a_1+a_2+...+a_n+c=k$.
By bars and stars theorem:
Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\binom{n+k}{n}$. That $\binom{n+k}{k} = \binom{n+k}{n}$.
but i don't know how to prove for negative integers.
 A: We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$. 
There will be $j\leq r:={\rm min}\{n,k\}$ of the $a_i$ that are $\ne0$, call them $a_{i_1}$, $\ldots$, $a_{i_j}$. The $i_l$ $(1\leq l\leq j)$ can be  chosen in ${n\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\geq 0$. Then we have to count the number of solutions to $x_1+\ldots+x_j+c=k-j$. This number is ${k\choose j}$. It has to be multiplied by $2^j$ in order to account for the  choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to
$$\sum_{j=0}^r{n\choose j}{k\choose j}2^j\ ,$$
which is symmetric in $n$ and $k$.
A: Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$. Note that
$$
f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0),\tag{*}
$$
by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get
$$
\begin{align}
f(n,k) 
&= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]\\
&= f(n-1,k)+f(n-1,k-1)+f(n,k-1)
\end{align}
$$
The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.
A: Introducing notation $A_i=|a_i|$ the first of inequalities can be rewritten as:
$$
A_1+A_2+...+A_k\le n.\tag1
$$
Any solution to $(1)$ can be viewed as composed of $i$ zeros ($0\le i\le k$) and $k-i$ non-zeros. Each non-zero value doubles the number of solutions to original equation (which can be both positive and negative). The overall number of solutions is therefore
$$
\sum_{i=0}^{k}\binom{k}{i}\binom{n}{k-i}2^{k-i}
\stackrel{i=k-j}=\sum_{j=0}^{k}\binom{k}{j}\binom{n}{j}2^{j}
=\sum_{j=0}^{\min(k,n)}\binom{k}{j}\binom{n}{j}2^{j},\tag2
$$
where $\binom{k}{i}$  counts the number of ways to place $i$ zeros into $k$ cells, and $\binom{n}{k-i}$ counts the number of positive solutions to the inequality
$$
p_1+p_2+\cdots+p_{k-i}\le n.
$$
In view of symmetry of $(2)$ with respect to interchange of $n$ and $k$ the original statement is proved.  
A: Just to show another approach to the answer already given by C.Blatter, let's start and consider 
$$
\left\{ \matrix{
  0 \le b_{\,j}  \in \mathbb Z \hfill \cr 
  b_{\,1}  + b_2  + b_{\,3}  +  \cdots  + b_{\,k}  \le n \hfill \cr}  \right.
$$
Then the number of solutions is the integral volume $V(k,n)$ of k-D simplex, with edge segments $(0,n)$, thus of length $n+1$.
It is not difficult to demonstrate that 
$$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k}=V(n,k)$$
then the hypothesis is already true in this case.
We can then partition the above solutions into those that contains $j$ variables
null and the remaining $k-j$ having values greater than $0$. We can assign the null variables in $\binom{k}{j}$ ways.
That corresponds to
$$
V(k,n)= \binom{n+k}{n}= \binom{n+k}{k}
 = \sum\limits_{0\, \le \,j\, \le \,k} {
 \binom{n}{k-j}  \binom{k}{j}
 } 
$$
Coming to our present case, for each non null variable $b_j$ we can assign two values to
the corresponding $a_j$, and just one to the null variables. So in this case the volume will be
$$
V_{\,a} \,(k,n) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\binom{n}{k-j} \binom{k}{j} 2^{\,k - j} } 
 = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\binom{n}{l} \binom{k}{l} 2^l }
 = V_{\,a} \,(n,k) 
$$
Therefore the hypothesis is demonstrated.
A: I just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is
$$
\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}
$$
which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction.
