About a combinatorics problem I'm dealing with this problem. Show that $\sum\limits_{j = 1}^n {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}i\\j\end{array}} \right)} \left( {\begin{array}{*{20}{c}}j\\
k\end{array}} \right) = 0$ for $1 \le i,k \le n$ and $i \ne k$.
Attempt: Since 
$\left( {\begin{array}{*{20}{c}}i\\j
\end{array}} \right) = 0$ for 
$j > i$ and 
$\left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right) = 0$ for 
$j < k$, it can be written as 
$\sum\limits_{j = k}^i {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}
i\\
j
\end{array}} \right)} \left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right)$. Also, we can let 
$i > k$ and so 
$i = k + a$ for some 
$1 \le a \le n$. But I can not continue.

The solution for odd values of $i + k$ is here. 
\begin{equation}
\sum\limits_{j = k}^i {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}
i\\
j
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right) = \sum\limits_{a = 0}^{i - k} {{{\left( { - 1} \right)}^a}\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right)} } . 
\end{equation}
Since $i + k$ is odd, $i-k$ is odd. Then, there are even number terms in this sum. Observe that 
\begin{equation}
\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right) = \frac{{i!}}{{\left( {i - \left( {k + a} \right)} \right)!\left( {k + a} \right)!}}.\frac{{\left( {k + a} \right)!}}{{k!a!}}
\end{equation}
\begin{equation}
 = \frac{{i!}}{{a!}}.\frac{1}{{\left( {i - \left( {k + a} \right)} \right)!k!}}
\end{equation}
Multiplying and dividing by $\left( {i - a} \right)!$, we get 
\begin{equation}
\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
i\\
{i - a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{i - a}\\
k
\end{array}} \right).
\end{equation}
Observe that the sign of
${\left( { - 1} \right)^a}$ is different from the sign of 
${\left( { - 1} \right)^{i - k - a}} = {\left( { - 1} \right)^{a + 1}}$ since 
$i-a = k + \left( {i - k - a} \right)$. Hence, this sum consists of equal terms having opposite sign and so is equal to $0$.
