$\binom {m}{0}+\binom {m}{1}-\binom {m}{2}-\binom {m}{3}+\binom {m}{4}+\binom {m}{5}-\binom {m}{6}-\binom {m}{7}+........=0$ if and only if for some positive integer k,then $m=$
$(A)4k\hspace{1cm} (B)4k+1\hspace{1cm}(C)4k+2\hspace{1cm}(D)4k+3$
My try:$\binom {m}{0}+\binom {m}{1}+\binom {m}{4}+\binom {m}{5}+....=\binom {m}{2}+\binom {m}{3}+\binom {m}{6}+\binom {m}{7}+.....$
But i could not judge the value of $m$.Any hints or guidance is required.
This falls into the following class of problems.
Evaluate $$S=\sum_{k=0}^n c_k\binom n k,$$ where the coefficients $c_k$ form a sequence that repeats periodically. That is, there exists an integer $m>0$ such that $c_{k+m}=c_k$ for all relevant $k$.
These can be handled as follows. Let $\zeta=e^{2\pi i/m}$. Then, by the binomial theorem, we have for all integers $\ell, 0\le\ell<m,$ the result $$ S(\ell)=\sum_{k=0}^n \zeta^{\ell k}\binom n k=(1+\zeta^\ell)^n. $$ Because $\zeta^{\ell(k+m)}=\zeta^{\ell k}\cdot\zeta^{m\ell}=\zeta^{\ell k},$ the sum $S(\ell)$ is of the above type. Furthermore, all the sums of this type are linear combinations of the sums $S(\ell)$. This is immediate from discrete Fourier analysis, but you can also see it as follows. The sequence of coefficients is fully determined, if we know $c_0,c_1,\ldots,c_{m-1}$. So if we can find coefficients $x_0,x_1,\ldots,x_{m-1}$ such that the system of linear equations $$ \begin{cases} 1\cdot x_0+1\cdot x_1+1\cdot x_2+\cdots+1\cdot x_{m-1}=c_0,&\\ 1\cdot x_0+\zeta\cdot x_1+\zeta^2\cdot x_2+\cdots+\zeta^{m-1}\cdot x_{m-1}=c_1,\\ \vdots\\ 1\cdot x_0+\zeta^j\cdot x_1+\zeta^{2j}\cdot x_2+\cdots+\zeta^{(m-1)j}\cdot x_{m-1}=c_j,\\ \vdots\\ 1\cdot x_0+\zeta^{m-1}\cdot x_1+\zeta^{2(m-1)}\cdot x_2+\cdots+\zeta^{(m-1)^2}\cdot x_{m-1}=c_{m-1} \end{cases} $$ holds, then $$ S=x_0 S(0)+x_1 S(1)+\cdots+x_{m-1}S(m-1). $$ You see that the determinant of the coefficient matrix of this system is of the Vandermonde type, so the system has a unique solution. Alternatively, you can do an inverse Fourier transform, and figure out that $$ x_\ell=\frac1m\sum_{j=0}^m\zeta^{-j\ell}c_j $$ is a solution.
In your case $m=4$, $c_0=c_1=1,$ $c_2=c_3=-1$ and $\zeta=i$. Leaving the rest to you.
The question is multiple-choice: to answer efficiently (and quickly) such questions when you are stuck proving the statement from scratch (or don't have time), you just have to eliminate 3 options to find the right (remaining) one. That is, find a counterexample for 3 out of 4 options, the simplest being by trying the simplest solution: here, for $k=1$.
The number $m $ should be odd.
because we should to show that $ \binom {m}{n} $ & $\binom {m}{m-n} $ should be be simplify with each other.
then $D$ is correct.
you can use induction that every number like $4k+3$ right for this equation.
