$\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26}=\frac{2}{3}(2^{27}+1)$ I need to prove the following identity 
$$
\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26} = \frac{2}{3}(2^{27}+1).
$$
I have tried to use the fact that $\binom{m}{n}=\binom{m}{m-n}$ but it doesn't help.

Another solution(just for fun)
Let $F_j(x)= \sum_{x \in 0..n}\binom{n}{x} | x \mod 3 = j$
$F_0(x) + F_1(x) +F_2(x) =2^x$
$F_j(x)=F_j(x-1) + F_{(j-1) \mod 3}(x-1)$
Self-cast previous equation
$F_j(x) = F_j(x-2) + 2*F_{(j-1) \mod 3}(x-2) + F_{(j-2) \mod 3}(x-2)$ =
 $2^{x-2} + F_{(j-1) \mod 3}(x-2)$
Another self-cast
$F_j(x) = 2^{x-2} + 2^{x-4} + 2^{x-6} + F_j(x-6)$
So, $F_2(28) = 2^{26} + 2^{24} + ... + 2^4 + F_2(4)$ = $2^{26} + 2^{24} + ... + 2^2 + F_1(2) = 2^{26} + 2^{24} + ... + 2^2 + \binom{2}{1} $ =
$2 + 4\sum_{i\in 0..12}4^i = 2 +4\frac{4^{13}-1}{3} = \frac{4^{14} - 4 + 6}{3}$ = $\frac{2^{28} +2}{3}$
