Problem - 3859
Show that $\binom{n}{0}+\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n$.
This is a basic identity which can be proved using binomial expansion. Because
$$(1+x)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} +\binom{n}{2} x^{n-2} +\cdots + \binom{n}{n-1}x+\binom{n}{n}$$
Setting $x=1$ to the above equation immediately leads to the to-be-claimed identity.