Binomial summation including de moivre's theorem

Question

Here i m also attaching the link to the solution if someone can refer through it and explain it to me it would be so helpful The question is simply asking $\sum_{r=0}^5 C_{6r}^{32}$. As I saw a jump of 6 in the binomial coefficients, I started thinking with the direction of de moivre's theorem and considered the expansion of $({1+x})^{32}$ and i m substituting the value of x by relating the de moivre's theorem using cos (something) + i sin (something) but i cannot get the correct approach and solve it

Answer 1

Just diving into the solution a bit, the first thing to recall is the binomial theorem: $$\begin{align} (x + y)^n &= \sum_{r=0}^n C_r^n x^{n-r} y^r \\ \Rightarrow (1 + x)^n &= \sum_{r=0}^n C_r^n x^{r} \end{align}$$ Recognizing that there is a "jump of 6" in evaluating $\sum_{r=0}^{5} C_{6r}^{32}$, the key is to then use the sixth roots of unity as given in the book: $$ x = \cos\frac{2\pi r}{6} + i \sin\frac{2\pi r}{6} = e^{2\pi i r/6} \quad ; \quad r = 0,1,\dots,5$$ Where de Moivre's theorem comes in is if we let $\xi = \cos\frac{2\pi}{6} + i \sin\frac{2\pi}{6} = e^{2\pi i/6}$, then the above values of $x$ are just $1, \xi, \xi^2, \dots \xi^5$. Then consider, $$\begin{align} (1 + 1)^{32} &= \sum_{r=0}^n C_r^n \\ (1 + \xi)^{32} &= \sum_{r=0}^n C_r^n \xi^{r}\\ (1 + \xi^2)^{32} &= \sum_{r=0}^n C_r^n \xi^{2r}\\ &\vdots\\ (1 + \xi^5)^{32} &= \sum_{r=0}^n C_r^n \xi^{5r} \end{align}$$ Adding these sums together gives, $$\begin{align} (1 + 1)^{32} + (1 + \xi)^{32} + \dots + (1 + \xi^5)^{32} &= \sum_{r=0}^n C_r^n \left(1 + \xi^{r} + \xi^{2r} + \dots + \xi^{5r} \right) \end{align}$$ Now notice that if $r$ is a multiple of $6$, then $\xi^r = 1$ so that $\left(1 + \xi^{r} + \xi^{2r} + \dots + \xi^{5r} \right) = 6$. Otherwise, we can use the geometric formula $$ 1 + \xi^{r} + \xi^{2r} + \dots + \xi^{5r} = \frac{1 - \xi^{r(5+1)}}{1 - \xi^r} = \frac{1 - 1}{1 - \xi^r} = 0$$ Thus, only every sixth term of the sum survives so that $$ (1 + 1)^{32} + (1 + \xi)^{32} + \dots + (1 + \xi^5)^{32} = 6 \sum_{r=0}^n C_{6r}^n $$ The solution in the book carries on from here.

Answer 2

let :$$w = \cos \left({\frac{\pi}{3}}\right)+i\sin\left({\frac{\pi}{3}}\right) \\ f(x)=(1+x)^n$$

We have : $$1+w^k+w^{2k}+w^{3k}+w^{4k}+w^{5k}=\begin{cases} 6|k : 6 \\ \text{otherwise :0}\end{cases}$$

we have :$$(1+x)^n=\sum^n_{k=0}\binom{n}{k}x^k$$ Therfore: $$f(1)+f(w)+f(w^2)+f(w^3)+f(w^4)+f(w^5)=\sum_{k=0}^{n}\binom{n}{k}\left({1+w^k+w^{2k}+w^{3k}+w^{4k}+w^{5k}}\right)=\sum_{k=0}^{\lfloor \frac{n}{6} \rfloor} 6\binom{n}{6k}$$

finally: $$\sum_{k=0}^{\lfloor \frac{n}{6} \rfloor} \binom{n}{6k}=\frac{f(1)+f(w)+f(w^2)+f(w^3)+f(w^4)+f(w^5)}{6}$$

Therfore: $$f(x)=(1+x)^{32}$$ $$\sum_{k=0}^{\lfloor \frac{32}{6} \rfloor} \binom{32}{6k}=\sum_{k=0}^{5} \binom{n}{6k}=\frac{f(1)+f(w)+f(w^2)+f(w^3)+f(w^4)+f(w^5)}{6}$$