Identity on Theory of Numbers

Question

Can anyone help me identify this identity? Or is there a known principle regarding this?

$k\binom{k}{k}-(k-1)\binom{k}{1}+(k-2)\binom{k}{2}-(k-3)\binom{k}{3}+\ldots +(-1)^{k-1}\binom{k}{k-1}$

Any help is highly appreciated.

Answer 1

This sum is nothing but $$S=\sum_{j=0}^{k} (-1)^j (k-j){k \choose j}$$ Change $j$ to $k-j$ to get $$S=\sum_{j=0}^{k} (-1)^{k-j} j{k \choose j}= (-1)^k \sum_{j=0}^{k} (-1)^j~ j~{k \choose j}~~~~(1)$$ Next D. w.r.t. $x$ the foloowing Binomial identity $$(1-x)^k= \sum (-1)^j {k \choose j} x^j.$$ $$k(1-x)^{k-1}=\sum_{j=0}^{k} (-1)^j ~j~ {k \choose j} x^{j-1}$$ By putting $x=1$ in above we get $$\sum_{j=0}^{k} (-1)^j~{k \choose j}=0$$ Hence, it follows that $S=0$.