I've found only this formula: $\sum_{j=k}^{n} \binom{n}{j} (-1)^j = (-1)^k \binom{n-1}{k-1}$;
But it seems useless. Please help.
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You can use $\sum_0^m = \sum_0^n - \sum_{m+1}^n$. – Vishu Dec 18 '22 at 19:12
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The identity $\binom nk = \binom n{n-k}$ would also work – Haydn Gwyn Dec 18 '22 at 19:14