<s>[INST] $\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26}=\frac{2}{3}(2^{27}+1)$ 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.
 [/INST] Hint. Let $w=\exp(2\pi i/3)$. Then $w^3=1$ and $1+w+w^2=0$, and hence
$$
(1+1)^{28}+w(1+w)^{28}+w^2(1+w^2)^{28}=3 \left(\binom{28}{2}+\binom{28}{5}+\cdots+\binom{28}{26}\right),
$$
since the sum above is a real number.
Next
$$
1+w=-w^2\quad\Rightarrow\quad(1+w)^{28}=w^{54}=w^2\\
1+w^2=-w\quad\Rightarrow\quad(1+w^2)^{28}=w^{28}=w
$$
Finally
$$
(1+1)^{28}+w(1+w)^{28}+w^2(1+w^2)^{28}=2^{28}+2.
$$
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