problem of binomial coefficients 
Find the value of $$\sum_{r=0}^5\binom{32}{6r}$$

Now I know this is a very easy question, but unfortunately I'm not getting the right answer.
I considered expansions of $(1+1)^{32},(1+\omega)^{32},(1+\omega^2)^{32},(1+\omega^3)^{32},(1+\omega^4)^{32},(1+\omega^5)^{32}$
After adding all of the above mentioned terms and dividing the result by $6$ should get us the required answer.
But when I am performing the above mentioned method, I am getting $$\frac{2^{33}-2}{6}$$ as answer which does not matches that of wolframaplha.
Any help is greatly appreciated.
 A: Let $\omega=\exp\big(i\frac{\pi}3\big)\cdot$
Observe that, for any integer $k$ we have $\sum_{n=0}^5 \omega^{nk}=
\begin{cases}
\frac{\omega^{6k-1}}{\omega^k-1}=0&\text{if}\ k\notin6\mathbb{Z},\\[3pt]
6&\text{if}\ k\in6\mathbb{Z}.
\end{cases}$
From the Newton binomial formula
\begin{eqnarray*}
\sum_{n=0}^5(1+\omega^n)^{32}
&=&\sum_{n=0}^5\sum_{k=0}^{32}\binom{32}{k}\omega^{nk}
=\sum_{k=0}^{32}\binom{32}{k}\sum_{n=0}^5\omega^{nk}\\
&=&\kern-0.7em\sum\limits_{k\in\{0,6,12,18,24,30\}}\kern-0.6em \binom{32}{k}\times6\\
&=&6\sum_{r=0}^5\binom{32}{6r}.
\end{eqnarray*}
So the required sum is $S=\big(\sum_{n=0}^5(1+\omega^n)^{32}\big)\big/6$.
Now we have $1+\omega^s
=\left\{
\begin{array}{rcl}
2&\text{if}\ s=0,\\[3pt]
\sqrt3\exp\big(i\frac\pi6\big)&\text{if}\ s=1,\\[3pt]
\exp\big(i\frac\pi3\big)&\text{if}\ s=2,\\[3pt]
0&\text{if}\ s=3,\\[3pt]
\exp\big({-}i\frac\pi3\big)&\text{if}\ s=4,\\[3pt]
\sqrt3\exp\big({-}i\frac\pi6\big)&\text{if}\ s=5.
\end{array}
\right.$
In the end we obtain
\begin{eqnarray*}
S&{}={}&\bigg[2^{32}
+3^{16}\!\exp\Big({-}i\frac{2\pi}3\Big)
+\exp\Big(i\frac{2\pi}3\Big)
+0
+\exp\Big({-}i\frac{2\pi}3\Big)
+3^{16}\!\exp\Big(i\frac{2\pi}3\Big)\bigg]\mathbin{\Big/}6\\
&{}={}&\frac{2^{32}-3^{16}-1}6\\
&{}={}&708653429.
\end{eqnarray*}
Edit (to answer a comment below).
I don't see any valuable generalization of the method above.
Suppose that one replaces 32 by $m$, 6 by $\lambda$ and puts $\ell=\lfloor m\lambda\rfloor$, in order to calculate $\sum_{n=0}^\ell\binom{m}{\lambda n}$.
We would take $\omega=\exp(2i\pi/\lambda)$, but the complication is that there's no reason in general to have $\sum\limits_{n\leqslant\ell}\omega^{kn}=0$ for $k\notin\lambda\mathbb{Z}$, as above.
In general this sum would be $\frac{\omega^{k(\ell+1)}-1}{\omega^k-1}\cdot$
With the values $(m,\lambda)=(32,6)$ given in the problem, everything went fine because $\ell+1\in\lambda\mathbb{Z}$ (i. e. $5+1\in6\mathbb{Z}$.
Perhaps the general $\displaystyle\sum_{n=0}^\ell\binom{m}{\lambda n}$ can be calculated by other means?
A: Having only 6 terms, I think the most straightforward way is to look at each coefficient individually. Let's call $S$ the desired sum.
$$S = \sum_{r=0}^5 \binom{32}{6r} = \binom{32}{0} + \binom{32}{6} + \binom{32}{12} + \binom{32}{18} + \binom{32}{24} + \binom{32}{30}$$
This shows that:
$$S = 1 + \dfrac{32!}{26!\cdot6!} + \dfrac{32!}{20!\cdot12!} + \dfrac{32!}{14!\cdot18!} + \dfrac{32!}{8!\cdot24!} + \dfrac{32!}{2!\cdot30!}$$
If we take $\dfrac{32!}{26!\cdot6!}$, it equals $\dfrac{32\cdot31\cdot30\cdot29\cdot28\cdot27}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 906192$.
Repeating the same process, we get $\binom{32}{12} = 225792840$, $\binom{32}{18} = 471435600$, $\binom{32}{24} = 10518300$, $\binom{32}{30} = \dfrac{32\cdot31}{2} = 16\cdot 31 = 496$.
Finally, $$\sum_{r=0}^5 \binom{32}{6r} = 1+906192+225792840+471435600+10518300+496 = 708653429$$
But all of this should not be done by hand, and has not been, it should instead be done with a computer, and seeing the prime factorization of the result, I really don't see how to bypass the steps mentioned above.
