Is there a formula for this sum:
$$ \sum_{j=0}^k {n \choose j} {n \choose k-j} (-2)^j \left(-\frac13 \right)^{k-j} ?$$
It reminds me to Vandermonde's identity; but as you can see there is a slight difference. I deduced this equivalent expression
$$ \left(- \frac13 \right)^k\sum_{j=0}^k {n \choose j} {n \choose k-j} (-2)^j \left(-3 \right)^{j}= \left(- \frac13 \right)^k\sum_{j=0}^k {n \choose j} {n \choose k-j} 6^{j},$$ but it doesn't seems to help so much.
Any help would be appreciated.