Bino-Combi-Turials! 2

We have $\begin{array}{rcl} S & = & \displaystyle \sum_{k=0}^{10}\binom{30}{3k} \; = \; \tfrac13\sum_{k=0}^{30} \binom{30}{k}(1 +\omega^k + \omega^{2k}) \\ & = & \displaystyle \tfrac13\big[2^{30} + (1+\omega)^{30} + (1 + \omega^2)^{30}\big] \; = \; \tfrac13\big[2^{30} + (-\omega^2)^{30} + (-\omega)^{30}\big] \; = \; \tfrac13(2^{30} +2) \end{array}$ making the answer the rather strange $\tfrac13(2^{30} - 2) = \boxed{357913940.\dot{6}}$ .

Let $\large k = \frac{ {30 \choose 1} + {30 \choose 2 }+ {30 \choose 4 }+... + {30 \choose 29}}{2}$ .

Consider the expansion of $(\omega + \omega^2)^{30}$ . We can easily see that this equals 1, and when expanded becomes

$({30 \choose 0} + {30 \choose 3} +... + {30 \choose 30}) - (\frac{1}{2})({30 \choose 1} + {30 \choose 2 }+ {30 \choose 4 }+... + {30 \choose 29}) = S-k$ . Thus, $S-k = 1$ .

Also, we can easily see that $S + 2k = 2^{30}$ , since ${30 \choose 0} + {30 \choose 1} + {30 \choose 2} +... + {30 \choose 30} = 2^{30}$ . We then have two equations, and solving them gives us $S = 357913942$ . We just then have to subtract $\frac{4}{3}$ from it (why?) to get the answer as

$\boxed{357913940.66}$

$S=2+\dfrac{30*29*28}{1*2*3}*\left (2+\dfrac{27*26*25}{4*5*6}*\left(2+\dfrac{24*23*22}{7*8*9}*\left(2+\dfrac{21*20*19}{10*11*12}*\left(2+\dfrac{18*17*16}{13*14*15} \right)\right)\right)\right)\\ =357913942.\\ \therefore~S-4/3=\huge \color{#D61F06}{357913940.6666667}.$

write the binomial equation, use x=1,w,w^2 to get the ans it's (2^30-2)/3. i'll give a pictorial sol. later!