How would I prove $\lceil \frac{\binom{2^{n+1}}{2}}{2^{n+1}}\rceil = 2^n$ for all natural numbers n?

Question

Any leads on how to prove $\lceil \frac{\binom{2^{n+1}}{2}}{2^{n+1}}\rceil = 2^n$?

Answer 1

Note that $\binom{m}2=\frac{m(m-1)}2$.

So

$$\lceil\frac{\binom{2^{n+1}}{2}}{2^{n+1}}\rceil= \lceil \frac{2^{n+1}( 2^{n+1}-1 )}{2\cdot2^{n+1}}\rceil=\lceil \frac{2^{n+1}-1 }{2}\rceil= \lceil{2^n-\frac{1 }{2}}\rceil=2^n$$