Q. 5 The Binomial Theorem says that a... [FREE SOLUTION]

Chapter 1: Q. 5 (page 151)

The Binomial Theorem says that an expression of the form ${(a + b)}^{n}$ can be expanded to localid="1649785698964" $(\begin{matrix} n \\ 0 \end{matrix}) a^{n} b^{0} + (\begin{matrix} n \\ 1 \end{matrix}) a^{n - 1} b^{1} + (\begin{matrix} n \\ 2 \end{matrix}) a^{n - 1} b^{2} + 鈰� + (\begin{matrix} n \\ n \end{matrix}) x^{0} y^{n},$ where for any localid="1649783317676" $0 鈮� k 鈮� n,$ the symbol $({_{k}}^{n})$ is equal to $\frac{n!}{k! (n - k)!} .$ 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 expressionlocalid="1649783538047" ${(1 + \frac{1}{n})}^{n} .$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

The given expression of ${(a + b)}^{n}$ is localid="1649785705634" ${(a + b)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) a^{n} b^{0} + (\begin{matrix} n \\ 1 \end{matrix}) a^{n - 1} b^{1} + (\begin{matrix} n \\ 2 \end{matrix}) a^{n - 2} b^{2} + 鈰� + (\begin{matrix} n \\ n \end{matrix}) a^{0} b^{n} .$

The given expression that needs to expand is localid="1649785353432" ${(1 + \frac{1}{n})}^{n} .$

localid="1649784210044" $({_{k}}^{n}) = \frac{n!}{k! (n - k)!} . 0! = 1 .$

Step 2. Simplification.

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

${(1 + \frac{1}{n})}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) 1^{n} {(\frac{1}{n})}^{0} + (\begin{matrix} n \\ 1 \end{matrix}) 1^{n - 1} {(\frac{1}{n})}^{1} + (\begin{matrix} n \\ 2 \end{matrix}) 1^{n - 2} {(\frac{1}{n})}^{2} + 鈰� + (\begin{matrix} n \\ n \end{matrix}) 1^{0} {(\frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n} = (\frac{n!}{0! (n - 0)!}) {(\frac{1}{n})}^{0} + (\frac{n!}{1! (n - 1)!}) {(\frac{1}{n})}^{1} + (\frac{n!}{2! (n - 2)!}) {(\frac{1}{n})}^{2} + 鈰� + (\frac{n!}{n! (n - n)!}) {(\frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n} = 1 + (\frac{n!}{1! (n - 1)!}) {(\frac{1}{n})}^{1} + (\frac{n!}{2! (n - 2)!}) {(\frac{1}{n})}^{2} + 鈰� + {(\frac{1}{n})}^{n}$

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

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Simplification.

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