Using Binomial Theorem to prove the following $$\large\sum_{j=0}^n (-1)^j {n\choose j}={n\choose 0}-{n\choose 1}+.....+\pm{n\choose n}=0 $$
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 ? 
 A: HINT
$$(1-1)^n =  \large\sum_{j=0}^n  \binom{n}{j}(1)^{n-j}(-1)^{j} $$
A: Hint: $$\large\sum_{j=0}^n x^j {n\choose j}=(1+x)^n$$
Can you use that to answer your question about even vs odd $n$?
A: It is interesting to consider some simple numerical examples here to get a more  intuitive feel of the formula. A bar above the number indicates a negative sign.
$n=2$:
$$1\quad \bar{2}\quad 1$$
$n=3$:
$$1\quad \bar{3}\quad 3\quad \bar{1}$$
$n=4$:
$$1\quad \bar{4}\quad 6\quad \bar{4}\quad 1$$
$n=5$:
$$1\quad \bar{5}\quad 10\quad \overline{10}\quad 5\quad\bar{1}$$
From the above it is clear that:


*

*for odd $n$, a coefficient has the opposite sign of its "mirror image", i.e. $${n\choose k}=-{n\choose n-k}$$ (e.g. 1, -1; -3, 3) thus cancelling out pairwise.

*for even $n$, this does not occur as a coefficient has the same sign as its "mirror image" (e.g. 1, 1; -4, -4); however, the sum of coefficients in even positions is numerically equal to the sum of coefficients in odd positions, but is negative, thus the sums cancel out (e.g. 1+6+1=4+4, and (1+6+1)+(-4-4)=0)
Thus the formula holds for both odd and even $n$. 
