How to formulate $(1-x)^k(1+x)^{n-k}$ as a polynomial sum expression Let $n$ and $k$ be natural numbers with $0\le k\le n$. I would like to write
$$(1-x)^k(1+x)^{n-k}$$ in the form of an explicit polynomial as
$$\sum_{j=0}^nb_j(n,k)x^j$$ Is there a way to write the integer functions
$b_j(n,k)$ as an expression hopefully with the help of binomials - perhaps as a sum of them ?
20150921 edited and added:
I hope to have correctly expanded and collected the double sum obtained using the binomial theorem on both factors. I obtained
$$\sum_{j=0}^n(-1)^{j}\left(\sum_{l=0}^j(-1)^l\binom{n-k}{l}\binom{k}{j-l}\right)x^j$$
The sum in the large parentheses reminds and is similar to the Vandermonde identity $$\sum_{l=0}^j\binom{x}{l}\binom{y}{j-l}=\binom{x+y}{l}$$
So my problem is now whether $$\sum_{l=0}^j(-1)^j\binom{x}{l}\binom{y}{j-l}=?$$ has a similar simplification (as a function probably) with $1$ binomial ? 
Additional comment
Perhaps at least a solution can be found/is possible in the special case that $x=n-k$ and $y=k$.
 A: Let $$p_{n,k}(x)=(1-x)^k(1+x)^{n-k}.$$
We easily see that $p_{n,k}(x)=p_{n,n-k}(x)$, so $b_j(n,k)=(-1)^jb_j(n,n-k)$.
If $2k\leq n$, then we get:
$$p_{n,k}(x)=(1-x)^k(1+x)^{n-k} = (1-x^2)^k(1+x)^{n-2k}$$
So $$b_j(n,k) = \begin{cases}
\displaystyle{\sum_{i=0}^{\lfloor j/2\rfloor} (-1)^i\binom{k}{i}\binom{n-2k}{j-2i}}&2k\leq n\\
\displaystyle{\sum_{i=0}^{\lfloor j/2\rfloor} (-1)^{i+j}\binom{n-k}{i}\binom{2k-n}{j-2i}}&2k> n
\end{cases}$$
You can actually use the first line for all $k$, as long as you define $\binom{a}{b}$ for negative ${a}$, which turns out to be equal to $(-1)^b\binom{b-a+1}{b}$.
You can also notice that $t^np_j(n,k)(1/t) = (-1)^kp_j(n,k)(t)$, which means that that $b_j(n,k)=(-1)^k b_{n-j}(n,k)$. So we can restrict our computations to when $2j\leq n$.

Additional notes:
Note that $p_{n+1,k}(x) = (1+x)p_{n,k}(x) =(1-x) p_{n,k-1}(x)$.
That shows that $b_{j}(n,k)+b_{j-1}(n,k) = b_{j}(n,k-1) - b_{j-1}(n,k-1)$.
That shows you get a recursion: $$b_j(n,k) = b_j(n,k-1)-b_{j-1}(n,k-1)-b_{j-1}(n,k)$$
You obviously have $b_0(n,k)=1$ and $b_n(n,k)=(-1)^k$.
The equation also shows that $b_j(n+1,k) = b_j(n,k) + b_{j-1}(n,k)$.
Another recursion can be found by taking the derivative of $p_{n,k}$ to get:
$$p_{n,k}'(x)= (n-k)p_{n-1,k}(x)-kp_{n-1,k-1}(x)$$
when $n,k>0$.
So you get: $$(j+1)b_{j+1}(n,k) = (n-k)b_{j}(n-1,k) -kb_j(n-1,k-1).$$
Another approach might be to take the generating function:
$$\begin{align}F(x,y,z)&=
\sum_{k_1,k_2=0}^{\infty} p_{k_1+k_2,k_1}(x)y^{k_1}z^{k_2}\\ &= 
\sum_{k_1,k_2} ((1-x)y)^{k_1}((1+x)z)^{k_2}\\
&=\frac{1}{1-(1-x)y} \frac{1}{1-(1+x)z}
\end{align}$$
Now you are trying to find the coefficients if $x^jy^kz^{n-k}$ in that expression.
Rewrite it as:
$$\frac{1}{1-y}\frac{1}{1+x\frac{y}{1-y}}\frac{1}{1-z}\frac{1}{1-x\frac{z}{1-z}}$$
or:
$$\left(\sum_{i=0}^\infty (-1)^ix^{i}\frac{y^i}{(1-y)^{i+1}}\right)
\left(\sum_{i=0}^\infty x^{i}\frac{z^i}{(1-z)^{i+1}}\right)$$
So if $B_j(y,z)$ is the coefficient of $x^j$ in this, then:
$$\begin{align}
B_i(y,z) &= 
\frac{1}{(1-y)(1-z)}\sum_{i=0}^j \left(\frac{y}{y-1}\right)^i\left(\frac{z}{1-z}\right)^{j-i}\\
&= \frac{1}{(1-y)(1-z)} \frac{\left(\frac{y}{y-1}\right)^{j+1}-\left(\frac{z}{1-z}\right)^{j+1}}{\frac{y}{y-1}-\frac{z}{1-z}}\\
&=\frac{\left(\frac{y}{y-1}\right)^{j+1}-\left(\frac{z}{1-z}\right)^{j+1}}{2yz-y-z}
\end{align}$$
This will ultimately just give you the formula arrived at above by others far more easily.
That would see to mean that $(2yz-y-z)F(x,y,z)$ is fairly simple. Is it?
It is, because $2-(1-x)-(1+x)=0$, so when $n>k$ then $2p_{n,k}(x)-p_{n+1,k}(x) - p_{n,k+1}(x)=0$. (This gives us the relationship $b_j(n+1,k) = 2b_j(n,k)-b_{j}(n,k+1)$ when $n>k$.)
So the only terms of the form $x^jy^{k_1}z^{k_2}$ are when $k_1=0$ or $k_2=0$, and they have simple coefficients.

Now: $$\left(\frac{u}{1-u}\right)^{j+1} = \sum_{k=0}^\infty \binom{k}{j+1}u^k$$
So we see that $$\begin{align}(y+z-2yz)F(x,y,z) 
&= \sum_j x^j\sum_{k=0}^\infty \binom{k}{j+1}\left(z^k + (-1)^{j+1}y^{k}\right)\\
&= \sum_{k} z^k \sum_j \binom{k}{j+1}x^j - \sum_{k} y^k \sum_j \binom{k}{j+1}(-x)^j
\end{align}$$
And $\sum_{j=0}^{\infty}\binom{k}{j+1}u^j = \frac{(1+u)^{k}-1}{u}$.
So:
$$\begin{align}x(y+z-2yz)F(x,y,z)&=\sum_{k=0}^{\infty} \left(((1+x)z)^{k}-((1-x)y)^{k}-(z^k-y^k))\right)\\
&=\frac1{1-(1+x)z} + \frac1{1-(1-x)y} - \frac{1}{1-z}+\frac{1}{1-y}
\end{align}$$
A: The coefficient of $x^h$ in $(1-x)^k$ is $\binom{k}{h}(-1)^h$, and the coefficient of $x^{j-h}$ in $(1+x)^{n-k}$ is $\binom{n-k}{j-h}$.
Therefore, taking care to get the limits of summation correct,
$$
b_j(n,k) \;= \sum_{h\,=\,\max(0,\,j+k-n)}^{\min(j,\,k)} \binom{k}{h} \binom{n-k}{j-h} (-1)^h .
$$
A: Note: A structural comparison of OPs binomial expression with the Vandermonde Identity
suggests that a representation without sums is not to expect.

If we set in OPs representation of Vandermonde's Identity $x=k$ and $y=n-k$ we obtain
  \begin{align*}
\sum_{l=0}^j\binom{k}{l}\binom{n-k}{j-l}=\binom{n}{j}\qquad\qquad 0\leq j,k\leq n\tag{1}
\end{align*}
  Since $\binom{n}{j}$ is the coefficient of $u^j$ in $(1+u)^n$ this expression corresponds with the algebraic representation
  \begin{align*}
(1+u)^k(1+u)^{n-k}=(1+u)^n\tag{2}
\end{align*}
  Let's revisit a proof of (1) using the coefficient of operator $[u^j]$ to denote the coefficient of $u^j$ in a polynomial $p(u)$. The left hand side gives
  \begin{align*}
[u^j]&(1+u)^k(1+u)^{n-k}\\
&=[u^j]\sum_{l=0}^{k}\binom{k}{l}u^l(1+u)^{n-k}\\
&=\sum_{l=0}^{k}\binom{k}{l}[u^{j-l}](1+u)^{n-k}\tag{3}\\
&=\sum_{l=0}^{k}\binom{k}{l}[u^{j-l}]\sum_{m=0}^{n-k}\binom{n-k}{m}u^m\\
&=\sum_{l=0}^{j}\binom{k}{l}\binom{n-k}{j-l}
\end{align*}
  and the right hand side 
  \begin{align*}
[u^j]&(1+u)^n=[u^j]\sum_{l=0}^{n}\binom{n}{l}u^l=\binom{n}{j}
\end{align*}

Comment: In (3) we use $[u^j]u^lp(u)=[u^{j-l}]p(u)$ and the linearity of the coefficient of operator is used in nearly each line. In the last line of the LHS the upper limit of the index is set to $j$. We follow thereby the convention that a binomial coefficient $\binom{n}{j}$ is set to zero if $j$ is less than zero or greater than $n$.

In (2) we see the LHS is the product of two polynomials which can be simplified to one polynomial $(1+u)^n$. On the other hand we cannot do a corresponding simplification with the expression.
\begin{align*}
(1-u)^k(1+u)^{n-k}=\sum_{j=0}^{n}b_j(n,k)u^j\tag{4}
\end{align*}
Since a binomial coefficient $\binom{r}{s}$ is the coefficient of $[u^s]$ of a polynomial $(1+u)^r$ it is not plausible that a simplification is possible.

$$ $$

A confirmation of OPs calculation: If we extract the coefficient of OP's polynomial (4) similar to above we obtain
\begin{align*}
[u^j]&(1-u)^k(1+u)^{n-k}\\
&=[u^j]\sum_{l=0}^{k}\binom{k}{l}(-u)^l(1+u)^{n-k}\\
&=\sum_{l=0}^{k}\binom{k}{l}(-1)^l[u^{j-l}](1+u)^{n-k}\\
&=\sum_{l=0}^{k}\binom{k}{l}(-1)^l[u^{j-l}]\sum_{m=0}^{n-k}\binom{n-k}{m}u^m\\
&=\sum_{l=0}^{j}(-1)^l\binom{k}{l}\binom{n-k}{j-l}\tag{5}
\end{align*}
  The expression (5) is the same as OPs result since
  \begin{align*}
\sum_{j=0}^n&(-1)^{j}\left(\sum_{l=0}^j(-1)^l\binom{n-k}{l}\binom{k}{j-l}\right)x^j\\
&=\sum_{j=0}^n(-1)^{j}\left(\sum_{l=0}^j(-1)^{j-l}\binom{n-k}{j-l}\binom{k}{l}\right)x^j\tag{6}\\
&=\sum_{j=0}^n\left(\sum_{l=0}^{j}(-1)^l\binom{k}{l}\binom{n-k}{j-l}\right)x^j\\
\end{align*}

Comment: In (6) we replace $l$ with $j-l$.
