Identity proof with binomial coefficients

Question

I have to prove the following identity and have to idea how to start.

$ { {-n} \choose k} = (-1)^k {{n+k-1} \choose k} $

I know that $ \sum_{k=0}^{n} (-1)^k{n \choose k}=0 $

but I cannot see a way, in which it could help prove the mentioned above.

Thanks for your help

Does this answer your question? Negative binomial coefficient — delivery101, Jul 02 '20 at 10:39
@NickyLevering partially. At the place the person in the thread writes:"the right side is clearly..." — Ozk, Jul 02 '20 at 11:16
@NickyLevering I do not really understand how he came to this transformation, would you mind explaining that? — Ozk, Jul 02 '20 at 11:16
We have: $\frac{n(n+1) \cdots (n+k-1)}{k!} = \frac{1}{k!}(n+k-1)(n+k-2) \cdots n = \frac{1}{k!}\frac{(n+k-1)!}{(n-1)!} = \binom{n+k-1}{k}$. — delivery101, Jul 02 '20 at 11:43

score 2 · Accepted Answer · answered Jul 02 '20 at 10:42

2

We can formally write $\binom {-n}{k}$ with factorial formula: $$ \binom {-n}{k}=\frac {(-n)(-n-1)(-n-2)\dotsm (-n-k+1)}{k!} =\\ (-1)^{k}\frac {(k+n-1)(k+n-2)\dotsm (n)}{k!}=(-1)^{k}\binom {k+n-1}{k} $$

answered Jul 02 '20 at 10:42

Vasily Mitch

10,414

Identity proof with binomial coefficients

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