Answer
$1$
Work Step by Step
We have to prove the identity:
$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^n\binom{n}{n}=0$
We have:
$\left(\dfrac{1}{4}+\dfrac{3}{4}\right)^5=1^5$
$\left(\dfrac{1}{4}+\dfrac{3}{4}\right)^5=1$
Use the Binomial Theorem to expand the left side:
$\binom{5}{0}\left(\dfrac{1}{4}\right)^5\left(\dfrac{3}{4}\right)^0+\binom{5}{1}\left(\dfrac{1}{4}\right)^4\left(\dfrac{3}{4}\right)^1+\binom{5}{2}\left(\dfrac{1}{4}\right)^3\left(\dfrac{3}{4}\right)^2+\binom{5}{3}\left(\dfrac{1}{4}\right)^2\left(\dfrac{3}{4}\right)^3+\binom{5}{4}\left(\dfrac{1}{4}\right)^1\left(\dfrac{3}{4}\right)^4+\binom{5}{5}\left(\dfrac{1}{4}\right)^0\left(\dfrac{3}{4}\right)^5=1$
$\binom{5}{0}\left(\dfrac{1}{4}\right)^5+\binom{5}{1}\left(\dfrac{1}{4}\right)^4\left(\dfrac{3}{4}\right)^1+\binom{5}{2}\left(\dfrac{1}{4}\right)^3\left(\dfrac{3}{4}\right)^2+\binom{5}{3}\left(\dfrac{1}{4}\right)^2\left(\dfrac{3}{4}\right)^3+\binom{5}{4}\left(\dfrac{1}{4}\right)^1\left(\dfrac{3}{4}\right)^4+\binom{5}{5}\left(\dfrac{3}{4}\right)^5=1$
So the given expression equals 1.
