Question

Show that ifis a positive integer, then

$\left.1=\left(\begin{array}{l}n\\ 0\end{array}\right)<\left(\begin{array}{c}n\\ 1\end{array}\right)<\cdots <\left(\begin{array}{c}n\\ \mid \frac{n}{2}\end{array}\right]\right)=\left(\begin{array}{c}n\\ \frac{n}{2}\end{array}\right])\mathbf{>}\mathbf{\cdots }\mathbf{>}\left(\begin{array}{c}n\\ n-1\end{array}\right)\mathbf{>}\left(\begin{array}{l}n\\ n\end{array}\right)\mathbf{=}\mathbf{1}$

Short answer

It's proven that$⌊\frac{n}{2}⌋=n-\left[\frac{n}{2}\right]\Rightarrow \left(\left[\begin{array}{c}n\\ \frac{n}{2}\end{array}\right]\right)=\left(\left[\begin{array}{c}n\\ \frac{n}{2}\end{array}\right)\right.$

Step-by-step solution

To prove

$\left.1=\left(\begin{array}{l}n\\ 0\end{array}\right)<\left(\begin{array}{c}n\\ 1\end{array}\right)<\cdots <\left(\begin{array}{c}n\\ \frac{n}{2}\end{array}\right)\right)=\left(\begin{array}{c}n\\ \frac{n}{2}\end{array}\right)>>>\left(\begin{array}{c}n\\ n-1\end{array}\right)>\left(\begin{array}{l}n\\ n\end{array}\right)=1.$

For positive integer n , state the required prove

$1\le k\le \left[\frac{n}{2}\right.⌋\left(\begin{array}{l}n\\ k\end{array}\right)-\left(\begin{array}{c}n\\ k-1\end{array}\right)=\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\frac{n-k+1}{k}-1\right)=\left(\begin{array}{c}n\\ k-1\end{array}\right)\cdot \frac{n-2k+1}{k}>0$whereas forthe other range of values of k , for

$\left[\frac{n}{2}\right]\le k\le n-1,\left(\begin{array}{c}n\\ k+1\end{array}\right)-\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{l}n\\ k\end{array}\right)\cdot \left(\frac{n-k}{k+1}-1\right)=\left(\begin{array}{l}n\\ k\end{array}\right)\cdot \frac{n-2k-1}{k+1}<0.$.The simple observation that $\left.\left.⌊\frac{n}{2}\right]=n-\left[\frac{n}{2}\right]\Rightarrow \left(\begin{array}{c}n\\ \frac{n}{2}\end{array}\right]\right)=\left(\begin{array}{c}n\\ \left.\frac{n}{2}\right]\end{array}\right)$ completes the proof.