A devious little problem indeed! I am interested in where you found it. We in fact have a very nice formula:
$$\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 =x^2$$
It follows from:
$$F(x)=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 \\ =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}\big(x^2-2kx+k^2\big) \\ =x^2\color{red}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}}-2x\color{green}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k}+\color{blue}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k^2}$$
Now, we need to somehow show
$$\color{red}{\blacksquare}=1 \\ \color{green}{\blacksquare}=\color{blue}{\blacksquare}=0$$
These sums have been studied before:
$\color{red}{\blacksquare}$ : Alternating sum of binomial coefficients equal to $1$
Follows from doing a binomial expansion of $(1-1)^n$.
$\color{green}{\blacksquare}$ : Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$
Follows from the recurrence $k\binom{n}{k}=n\binom{n-1}{k-1}$.
$\color{blue}{\blacksquare}$ : This is the hard one. We proceed as follows:
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k \binom{n-1}{k-1} \tag{1}$$ $$ =n\sum_{l=0}^{n-1}(-1)^l(l+1)\binom{n-1}{l} \\ =n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}+n\underbrace{\sum_{l=0}^{n-1}(-1)^l \binom{n-1}{l}}_{=(1-1)^{n-1}=0} $$
Finally, since $\binom{n-1}{n}=0$, and since the $l=0$ summand is zero, we can remove the $l=0$ index and add a $l=n$ index, and then rename the index back to $k$:
$$n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}=n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}$$
Now, using the recurrence relation for the binomials,
$$n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}=n\underbrace{\sum_{k=1}^n k(-1)^k \binom{n}{k}}_{=\color{green}{\blacksquare}=0}-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}$$
Again, $k\binom{n-1}{k-1}=k^2\binom{n}{k}$ and hence
$$-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}=-n\sum_{k=1}^n (-1)^kk^2 \binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k^2 \binom{n}{k}\tag{2}$$
But, retracing our steps from $(1)$ to $(2)$, we have just proved that
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=~\boldsymbol{n}~\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}$$
This can only be true for general $n$ if the sum is zero. Hence,
$$\boxed{\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2=x^2\color{red}{\blacksquare}-2x\color{green}{\blacksquare}+\color{blue}{\blacksquare} \\ =x^2}$$
QED!!
I saw this question or a similar in this site but it is not easy to find it. With the hints in the comments:
By binomial expansion, we have
$$\begin{array}
.(1+1)^n&=&\binom{n}{0}&+\binom{n}{1}&+\binom{n}{2}&+\binom{n}{3}\cdots\\
(1+w)^n&=&\binom{n}{0}&+\binom{n}{1}w&+\binom{n}{2}w^2&+\binom{n}{3}w^3+\cdots\\
(1+w^2)^n&=&\binom{n}{0}&+\binom{n}{1}w^2&+\binom{n}{2}w^4&+\binom{n}{3}w^6+\cdots
\end{array}$$
Now, by using the identities $1+w+w^2=0$ and $w^3=1$, we can compute the linear combinations below. For example, $3a_n$ is just the sum of the equations above.
$$a_n=\frac{1}{3}\left((1+1)^n+(1+w)^n+(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3})\right)$$
$$b_n=\frac{1}{3}\left((1+1)^n+w^2(1+w)^n+w(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3}+\frac{4\pi}{3})\right)$$
$$c_n=\frac{1}{3}\left((1+1)^n+w(1+w)^n+w^2(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3}+\frac{2\pi}{3})\right)$$
Now you can check these by using the identities $\cos(u+v)=\cos u\cos v-\sin u\sin v$, $\cos(\frac{2\pi}{3})=\cos(\frac{4\pi}{3})=-\frac{1}{2}$, $\sin(\frac{2\pi}{3})=-\sin(\frac{4\pi}{3})=\frac{\sqrt{3}}{2}$ or some better identities you know:
$$\cos(\frac{n\pi}{3})+\cos(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos(\frac{n\pi}{3}+\frac{4\pi}{3})=0,$$
$$\cos^2(\frac{n\pi}{3})+\cos^2(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos^2(\frac{n\pi}{3}+\frac{4\pi}{3})=\frac{3}{2},$$
$$\cos(\frac{n\pi}{3})\cos(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos(\frac{n\pi}{3})\cos(\frac{n\pi}{3}+\frac{4\pi}{3})+\cos(\frac{n\pi}{3}+\frac{2\pi}{3})\cos(\frac{n\pi}{3}+\frac{4\pi}{3})=-\frac{3}{4}.$$
Hence, the sum is $\frac{4}{9}\left(\frac{3}{2}-\left(-\frac{3}{4}\right)\right)=1.$
This also immediately follows from achille hui's hint and these computations are not necessary:
$$x^2+y^2+z^2-(xy+yz+zx)=(x+yw+zw^2)(x+yw^2+w^4)$$
gives
$$a_n^2+b_n^2+c_n^2-(a_nb_n+a_nc_n+b_nc_n)=(a_n+b_nw+c_nw^2)(a_n+b_nw^2+c_nw^4)=(1+w)^n(1+w^2)^n=(1+w+w^2+w^3)^n=w^{3n}=1.$$
Best Answer
We will use $i=\sqrt{-1}$. Then define:
$$A=(1+1)^{200}=\sum_{k=0}^{200}{200 \choose k} 1^k$$
$$B=(1-1)^{200}=\sum_{k=0}^{200}{200 \choose k} (-1)^k$$
$$C=(1+i)^{200}=\sum_{k=0}^{200}{200 \choose k} i^k$$
$$D=(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (-i)^k$$
Now take A+B-C-D:
$$2^{200} + 0-(1+i)^{200}-(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (1+(-1)^k-i^k-(-i)^k)$$
Note that $(1+(-1)^k-i^k-(-i)^k)$ is equal to zero unless $k \equiv 2 \pmod 4$, when it is $4$.
$$2^{200} -2 \times 2^{100}=8\sum_{j=0}^{24}{200 \choose {4j+2}} $$
$$\sum_{j=0}^{24}{200 \choose {4j+2}} =2^{197} - 2^{98} $$