Q. 5 - Calculus

Question

The Binomial Theorem says that an expression of the form ${\left(a+b\right)}^n$can be expanded to $\left(\begin{array}{c}n\\ 0\end{array}\right)a^n{b}^0+\left(\begin{array}{c}n\\ 1\end{array}\right)a^{n-1}{b}^1+\left(\begin{array}{c}n\\ 2\end{array}\right)a^{n-1}{b}^2+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)x^0{y}^n,$where for any $0\le k\le n,$the symbol $\left({{}_k}^n\right)$is equal to $\frac{n!}{k!\left(n-k\right)!}.$Here $n!$is n factorial, the product of the integers from $1$to n. By convention we set $0!=1.$Apply this expansion to the expression ${\left(1+\frac{1}{n}\right)}^n.$

Step-by-Step Solution

Answer

 Expanded form of ${\left(1+\frac{1}{n}\right)}^n$is ${\left(1+\frac{1}{n}\right)}^n=1+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +{\left(\frac{1}{n}\right)}^n.$

Step 1. Given Information.

The given expression of ${\left(a+b\right)}^n$is ${\left(a+b\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)a^n{b}^0+\left(\begin{array}{c}n\\ 1\end{array}\right)a^{n-1}{b}^1+\left(\begin{array}{c}n\\ 2\end{array}\right)a^{n-2}{b}^2+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^0{b}^n.$

The given expression that needs to expand is ${\left(1+\frac{1}{n}\right)}^n.$

$$\begin{gathered} \left({{}_k}^n\right)=\frac{n!}{k!\left(n-k\right)!}. \\ 0!=1. \end{gathered}$$

Step 2. Simplification.

Expand the expression ${\left(1+\frac{1}{n}\right)}^n.$

$$\begin{gathered} {\left(1+\frac{1}{n}\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)1^n{\left(\frac{1}{n}\right)}^0+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}{\left(\frac{1}{n}\right)}^1+\left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}{\left(\frac{1}{n}\right)}^2+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)1^0{\left(\frac{1}{n}\right)}^n \\ {\left(1+\frac{1}{n}\right)}^n=\left(\frac{n!}{0!(n-0)!}\right){\left(\frac{1}{n}\right)}^0+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +\left(\frac{n!}{n!(n-n)!}\right){\left(\frac{1}{n}\right)}^n \\ {\left(1+\frac{1}{n}\right)}^n=1+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +{\left(\frac{1}{n}\right)}^n \end{gathered}$$

So the expanded form is ${\left(1+\frac{1}{n}\right)}^n=1+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +{\left(\frac{1}{n}\right)}^n.$