Chapter 12.9, Problem 27E

To determine

To proof: the given statement by using mathematical induction.

Explanation of Solution

Given information:

Binomial Theorem is valid for all positive integral values of $n$ .

Calculation:

Binomial Theorem states that for $n\ge 0$ ,

  ${\left(a+b\right)}^n={\displaystyle \sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)}{a}^{n-k}{b}^k$

The base for $n=0$ is as follows:

  $\begin{array}{c}{\left(a+b\right)}^0={\displaystyle \sum _{k=0}^0\left(\begin{array}{l}0\\ 0\end{array}\right)}{a}^0{b}^0\\ =1\end{array}$

The induction hypothesis for $n\ge 0$ is as follows:

  ${\left(a+b\right)}^n={\displaystyle \sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)}{a}^{n-k}{b}^k$

The induction step is given below.

We need to prove that for $n\ge 0$ ,

  $\begin{array}{l}{\left(a+b\right)}^{n+1}={\displaystyle \sum _{k=0}^{n+1}\left(\begin{array}{l}n+1\\ k\end{array}\right)}{a}^{n+1-k}{b}^k\\ \text{i}\text{.e}\text{. }{\left(a+b\right)}^{n+1}=\left(a+b\right){\left(a+b\right)}^n\end{array}$

By using the induction hypothesis,

  ${\left(a+b\right)}^{n+1}=\left(a+b\right){\displaystyle \sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)}{a}^{n-k}{b}^k$

  $=\left(a+b\right)\left[\left(\begin{array}{l}n\\ 0\end{array}\right)a^n+\left(\begin{array}{l}n\\ 1\end{array}\right)a^{n-1}b+\mathrm{.....}+\left(\begin{array}{l}n\\ n-1\end{array}\right)a{b}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)b^n\right]$

  $=\left[\left(\begin{array}{l}n\\ 0\end{array}\right)a^{n+1}+\left(\begin{array}{l}n\\ 1\end{array}\right)a^{n}b+\mathrm{.....}+\left(\begin{array}{l}n\\ n-1\end{array}\right)a^2{b}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)a{b}^n\right]+$

  $\left[\left(\begin{array}{l}n\\ 0\end{array}\right)a^{n}b+\left(\begin{array}{l}n\\ 1\end{array}\right)a^{n-1}{b}^2+\mathrm{....}+\left(\begin{array}{l}n\\ n-1\end{array}\right)a{b}^n+\left(\begin{array}{l}n\\ n\end{array}\right)b^{n+1}\right]$

  $=\left(\begin{array}{l}n\\ 0\end{array}\right)a^{n+1}+\left[\left(\begin{array}{l}n\\ 0\end{array}\right)+\left(\begin{array}{l}n\\ 1\end{array}\right)\right]a^{n}b+\mathrm{....}+\left[\left(\begin{array}{l}n\\ n-1\end{array}\right)+\left(\begin{array}{l}n\\ n\end{array}\right)\right]a{b}^n+\left(\begin{array}{l}n\\ n\end{array}\right)b^{n+1}\mathrm{...}\left(1\right)$

We have completed our proof with two identities for any natural number $k$ and for any natural number $k\text{ and }n$ as follows:

  $\begin{array}{l}\left(\begin{array}{l}k\\ 0\end{array}\right)=\left(\begin{array}{l}k\\ k\end{array}\right)=1\\ \left(\begin{array}{l}n\\ k\end{array}\right)+\left(\begin{array}{l}n\\ k+1\end{array}\right)=\left(\begin{array}{c}n+1\\ k+1\end{array}\right)\end{array}$

Thus equation (1) can be rewritten as follows:

  ${\left(a+b\right)}^{n+1}=\left(\begin{array}{l}n+1\\ 0\end{array}\right)a^{n+1}+\left[\left(\begin{array}{l}n+1\\ 1\end{array}\right)\right]a^{n}b+\mathrm{....}+\left[\left(\begin{array}{l}n+1\\ n\end{array}\right)\right]a{b}^n+\left(\begin{array}{l}n+1\\ n+1\end{array}\right)b^{n+1}$

Hence, the given condition is proved.