Finding $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}\cdots \cdots +(-1)^r\binom{n}{r}$
Try: using Identity $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$
So $$\sum^{r}_{k=1}(-1)^r\bigg(\binom{n-1}{k}+\binom{n-1}{k-1}\bigg)$$
Ciuld some help me how to find sum of that series, Thanks
Looks like it telescopes...to give you ${{n-1} \choose 0}+{ {n-1} \choose r }$, which is ${n-1} \choose r $...
I believe there's a typo: should be $(-1)^k $...
$$\sum_{k=0}^{r}{(-1)^k\binom{n}{k}}=(-1)^r\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}}=(-1)^r\sum_{k=0}^{n}{\binom{-1}{r-k}\binom{n}{k}}=(-1)^r\binom{n-1}{r}$$
This is somebody else's clever idea, but I can't find a link to it.
