Evaluation of the following algebraic equation using binomial coefficients

Question

Given \begin{align} a &= {2015\choose0}+{2015\choose3}+{2015\choose6}+\cdots \\[2pt] b &= {2015\choose1}+{2015\choose4}+{2015\choose7}+\cdots \\[2pt] c &= {2015\choose2}+{2015\choose5}+{2015\choose8}+\cdots \end{align} We have to find $(b-c)^2+(c-a)^2$

Note: Here $${n\choose{k}}=\frac{n!}{k!\cdot(n-k)!}$$ where $k \leq n$ for all $n,k \in \mathbb{N}$. It is called the binomial coefficient.

I have observed that $(a+b+c)=2^{2015}$ using the binomial expansion but the given fact does not provide clue to solve the given problem. Hence how can we solve the given problem?

Answer 1

Using Sum of Binomial Series of form $\binom{2000}{3k-1}$, since $\binom{n}{m} = \binom{n}{n-m}$, we first have

$$\begin{equation}\begin{aligned} c & = \binom{2015}{2} + \binom{2015}{5} + \binom{2015}{8} + \ldots + \binom{2015}{2015} \\ & = \binom{2015}{2015} + \binom{2015}{2012} + \ldots + \binom{2015}{5} + \binom{2015}{2} \\ & = \binom{2015}{0} + \binom{2015}{3} + \ldots + \binom{2015}{2010} + \binom{2015}{2013} \\ & = a \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Thus, from your result, we get

$$a + b + c = 2a + b = 2^{2015} \tag{2}\label{eq2A}$$

Similar to Feng's comment suggestion, and what's used in this answer, let $\omega = e^{(2\pi /3)i}$, so $\omega^3 = 1$. Then, by the binomial theorem,

$$(1+\omega)^{2015}=\binom{2015}{0}+\binom{2015}{1}\omega+\binom{2015}{2}\omega^2+\ldots+\binom{2015}{2015}\omega^2 \tag{3}\label{eq3A}$$

Comparing the real parts of both sides gives

$$\Re\left((1+\omega)^{2015}\right)=\Re\left(\binom{2015}{0}+\binom{2015}{1}\omega+\binom{2015}{2}\omega^2+\ldots+\binom{2015}{2015}\omega^2\right) \tag{4}\label{eq4A}$$

Next, note that

$$\begin{equation}\begin{aligned} 1 + \omega & = 1 + \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) \\ & = 1 - \frac{1}{2} + i\left(\frac{\sqrt{3}}{2}\right) \\ & = \frac{1 + \sqrt{3}i}{2} \\ & = e^{(\pi/3)i} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Since $(e^{(\pi/3)i})^3 = -1$ and $2015 = 671(3) + 2$, we then have

$$(1+\omega)^{2015} = -e^{(2\pi/3)i} = -\omega \tag{6}\label{eq6A}$$

Also, since $\cos\left(\frac{2\pi}{3}\right) = \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}$, we've got that

$$\Re(\omega) = \Re(\omega^2) = -\frac{1}{2} \tag{7}\label{eq7A}$$

Using \eqref{eq6A} and \eqref{eq7A} in \eqref{eq4A}, then

$$\frac{1}{2} = a - \frac{b}{2} - \frac{a}{2} = \frac{a}{2} - \frac{b}{2} \;\to\; a - b = 1 \tag{8}\label{eq8A}$$

Adding \eqref{eq2A} and \eqref{eq8A} gives

$$3a = 2^{2015} + 1 \;\;\to\;\; a = \frac{2^{2015} + 1}{3} \tag{9}\label{eq9A}$$

Next, \eqref{eq2A} minus $2$ times \eqref{eq8A} results in

$$3b = 2^{2015} - 2 \;\;\to\;\; b = \frac{2^{2015} - 2}{3} \tag{10}\label{eq10A}$$

Finally, using \eqref{eq1A}, \eqref{eq9A} and \eqref{eq10A}, we get

$$\begin{equation}\begin{aligned} (b - c)^2 + (c - a)^2 & = \left(\frac{2^{2015} - 2}{3} - \frac{2^{2015} + 1}{3}\right)^2 + \left(\frac{2^{2015} + 1}{3} - \frac{2^{2015} + 1}{3}\right)^2 \\ & = \left(\frac{-3}{3}\right)^2 + 0 \\ & = 1 \end{aligned}\end{equation}\tag{11}\label{eq11A}$$

Answer 2

This is along the general lines of Feng's comment and John's post, with just simpler algebra.

As you noted, $a+b+c=(1+1)^{2015} = 2^{2015}=n$, say. Using the non-real cube root of unity which satisfies $\omega^3=1$ and $1+\omega+\omega^2=0$, you can similarly note that
$a+b\omega+c\omega^2=(1+\omega)^{2015}=(-\omega^2)^{2015}=-\omega$ and $a+b\omega^2+c\omega=(1+\omega^2)^{2015}=(-\omega)^{2015}=-\omega^2$.

Adding those three, we get
$3a=n-\omega-\omega^2=n+1$. Similarly using the above three equations, we also get
$3b=n+(-\omega)\omega^2+(-\omega^2)\omega=n-2$ and
$3c=n+(-\omega)\omega+(-\omega^2)\omega^2=n-\omega^2-\omega=n+1$

Hence $3b-3c=-3$ and $3c-3a=0$, giving $(b-c)^2+(c-a)^2=(-1)^2+0=1.$