Chapter 12.6, Problem 51E

To determine

To Prove: $\left(\begin{array}{c}n\\ 1\end{array}\right)=n$ and $\left(\begin{array}{c}n\\ n-1\end{array}\right)=n$ using Binomial Theorem.

Explanation of Solution

Given information:

Using the Binomial Theorem.

Concept used:

According to the Binomial theorem, expansion of ${\left(x+y\right)}^n$ with non-negative integer $n$ is:

  ${\left(x+y\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n{y}^0+\left(\begin{array}{l}n\\ 1\end{array}\right)x^{n-1}{y}^1+\mathrm{.............}+\left(\begin{array}{l}n\\ n-1\end{array}\right)x^1{y}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)x^0{y}^n$

Where $n\ge 0$ , and $\left(\begin{array}{l}n\\ k\end{array}\right)$ denotes the binomial coefficient for $k=0,1,\mathrm{2........}n-1,n$

The formula for $\left(\begin{array}{l}n\\ k\end{array}\right)$ is,

  $\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{\left(n-k\right)!k!}$

The number of terms in expansion of ${\left(x+y\right)}^n$ is $n+1$

Proof:

  $\begin{array}{l}\left(\begin{array}{c}n\\ 1\end{array}\right)=\frac{n!}{\left(n-1\right)!1!}\\ \Rightarrow \left(\begin{array}{c}n\\ 1\end{array}\right)=\frac{n\times \left(n-1\right)!}{\left(n-1\right)!}\\ \Rightarrow \left(\begin{array}{c}n\\ 1\end{array}\right)=n\end{array}$

  $\begin{array}{l}\left(\begin{array}{c}n\\ n-1\end{array}\right)=\frac{n!}{\left(n-\left(n-1\right)\right)!\left(n-1\right)!}\\ \Rightarrow \left(\begin{array}{c}n\\ n-1\end{array}\right)=\frac{n\times \left(n-1\right)!}{\left(n-n+1\right)!\left(n-1\right)!}\\ \Rightarrow \left(\begin{array}{c}n\\ n-1\end{array}\right)=\frac{n}{1!}\\ \Rightarrow \left(\begin{array}{c}n\\ n-1\end{array}\right)=n\end{array}$

Hence proved.