With the Binomial Theorem

Question

How do I prove the following using the binomial theorem?

$$\sum_{j=0}^n (-1)^j{n \choose j} = {n \choose 0}-{n \choose 1}+\ldots\pm{n \choose n}= 0$$

Answer 1

Note that $(a+b)^n=\sum\limits_{j=0}^n{n\choose j}a^{n-j}b^j$. Substituting $a=1,b=-1$, we have $0=(1-1)^n=\sum\limits_{j=0}^n{n\choose j}1^{n-j}(-1)^j=\sum\limits_{j=0}^n(-1)^j{n\choose j}$