Question

Let $n$ be a positive integer. Prove that the binominal coefficients $\left(\begin{array}{c}n\\ 1\end{array}\right),\left(\begin{array}{c}n\\ 2\end{array}\right),\left(\begin{array}{c}n\\ 3\end{array}\right).........,\left(\begin{array}{c}n\\ n-1\end{array}\right)$ are all even if and only if $n$ is a power of $2$.

Answer 1

As we can clearly see $n\in$Odd Number ${\Rightarrow }^n{C}_1\in$ Odd Number.

So, $n=2^{m}k$ where $m\in$Integers, $k\in$ Odd Numbers.

Let us assume $n\ne 2^m$.

ie, $k>1$

Then, $\left({}_{2^m}^{2^{m}k}\right)=\frac{\left(2^{m}k\right)!}{(2^{m}k-2^m)!\left(2^m\right)!}=\frac{\left(2^{m}k\right)!}{\left(2^m\right(k-1\left)\right)!\left(2^m\right)!}=\frac{2^m(k-1)+1}{1}\frac{2^m(k-1)+2}{2}...\frac{2^m(k-1)+2^m}{2^m}$

As we can see here, $2^m(k-1)$ is an even number.

So, $2^m(k-1)+$ Odd Number is an Odd Number.

Also, $\frac{2^m(k-1)+E}{E}$, where $E$ is an even number is also an even number. ($\because$all the powers of $2$ will get cancelled out).

As every factor of the product is an odd number, the product itself is an odd number.

So, our assumption was wrong.

ie, $n=2^m$.