$\sum \binom{n}{r}$ from (r=0 to r=n) is equal to 2^n.
it includes both even and and odd terms . ( sum of odd terms + sum of even terms = 2^n).......(1)
and also sum of even terms = sum of odd terms..........(2).
solving eq.(1) and eq (2) we can say that sum of odd terms = 2^(n-1).
so $\sum_{r=1}^{13} \binom{13}{r}$ = 2^(13-1)= 4096.
but we need the sum from $\binom{13}{3} +\binom{13}{5} +.......\binom{13}{13}$
since 2^(13-1) includes terms from $ \binom{13}{1} + \binom{13}{3} +\binom{13}{5} +.......\binom{13}{13}$
. so just subtract $\binom{13}{1}$ from 4096.
==> 4096-13= 4083 ans.
