[Math] Using Binomial Theorem to prove the following

Question

This question already has answers here:

Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$

(7 answers)

Closed 7 years ago.

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

I'm confused by the last part of the equation $\pm$. it seems imply that the sum would be equal to 0 no matter the $n$ is even or odd ?

Answer 1

HINT

$$(1-1)^n = \large\sum_{j=0}^n \binom{n}{j}(1)^{n-j}(-1)^{j} $$