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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.

RobPratt
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rowcol
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    Not sure how helpful this is, but this should be equivalent to the coefficient of $x^k$ in the expansion of $(1+x)^n(1+6x)^n$, times $(-1/3)^k$.

    Edit: looking at https://math.stackexchange.com/questions/2008583/closed-form-for-weighted-sum-involving-product-of-binomial-coefficients, it looks like there might not be a nice closed form for expressions like this.

     – octave Mar 23 '21 at 19:34

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