To determine To find: To find the value of ( 5 0 ) ( 1 4 ) 5 + ( 5 1 ) ( 1 4 ) 4 ( 3 4 ) + ( 5 2 ) ( 1 4 ) 3 ( 3 4 ) 2 + ( 5 3 ) ( 1 4 ) 2 ( 3 4 ) 3 + ( 5 4 ) ( 1 4 ) 1 ( 3 4 ) 4 + ( 5 5 ) ( 3 4 ) 5 . Answer Solution: The value of as ( 5 0 ) ( 1 4 ) 5 + ( 5 1 ) ( 1 4 ) 4 ( 3 4 ) + ( 5 2 ) ( 1 4 ) 3 ( 3 4 ) 2 + ( 5 3 ) ( 1 4 ) 2 ( 3 4 ) 3 + ( 5 4 ) ( 1 4 ) 1 ( 3 4 ) 4 + ( 5 5 ) ( 3 4 ) 5 = 1 . Explanation Given: Expression is given as ( 5 0 ) ( 1 4 ) 5 + ( 5 1 ) ( 1 4 ) 4 ( 3 4 ) + ( 5 2 ) ( 1 4 ) 3 ( 3 4 ) 2 + ( 5 3 ) ( 1 4 ) 2 ( 3 4 ) 3 + ( 5 4 ) ( 1 4 ) 1 ( 3 4 ) 4 + ( 5 5 ) ( 3 4 ) 5 . Formula used: The Binomial Theorem: Let x and a be real numbers. For any positive integer n , we have ( x + a ) n = ( n 0 ) x n + ( n 1 ) a x n − 1 + ⋯ + ( n j ) a j x n − j + ⋯ + ( n n ) a n ( x + a ) n = Σ k = 0 n ( n k ) x n − j a j Here x = 1 4 , a = 3 4 and n = 5 . Applying Binomial theorem, ( 5 0 ) ( 1 4 ) 5 + ( 5 1 ) ( 1 4 ) 4 ( 3 4 ) + ( 5 2 ) ( 1 4 ) 3 ( 3 4 ) 2 + ( 5 3 ) ( 1 4 ) 2 ( 3 4 ) 3 + ( 5 4 ) ( 1 4 ) 1 ( 3 4 ) 4 + ( 5 5 ) ( 3 4 ) 5 = ( 1 4 + 3 4 ) 5 = ( 1 ) 5 = 1

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Textbook Question

Chapter 12.5, Problem 49AYU

$(\begin{array}{l} 5 \\ 0 \end{array}) {(\frac{1}{4})}^{5} + (\begin{array}{l} 5 \\ 1 \end{array}) {(\frac{1}{4})}^{4} (\frac{3}{4}) + (\begin{array}{l} 5 \\ 2 \end{array}) {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{2} + (\begin{array}{l} 5 \\ 3 \end{array}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3} + (\begin{array}{l} 5 \\ 4 \end{array}) (\frac{1}{4}) {(\frac{3}{4})}^{4} + (\begin{array}{l} 5 \\ 5 \end{array}) {(\frac{3}{4})}^{5} = ?$

Expert Solution & Answer

To determine

To find: To find the value of $(\begin{matrix} 5 \\ 0 \end{matrix}) {(\frac{1}{4})}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(\frac{1}{4})}^{4} (\frac{3}{4}) + (\begin{matrix} 5 \\ 2 \end{matrix}) {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{4} + (\begin{matrix} 5 \\ 5 \end{matrix}) {(\frac{3}{4})}^{5}$ .

Answer to Problem 49AYU

Solution:

The value of as $(\begin{matrix} 5 \\ 0 \end{matrix}) {(\frac{1}{4})}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(\frac{1}{4})}^{4} (\frac{3}{4}) + (\begin{matrix} 5 \\ 2 \end{matrix}) {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{4} + (\begin{matrix} 5 \\ 5 \end{matrix}) {(\frac{3}{4})}^{5} = 1$ .

Explanation of Solution

Given:

Expression is given as $(\begin{matrix} 5 \\ 0 \end{matrix}) {(\frac{1}{4})}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(\frac{1}{4})}^{4} (\frac{3}{4}) + (\begin{matrix} 5 \\ 2 \end{matrix}) {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{4} + (\begin{matrix} 5 \\ 5 \end{matrix}) {(\frac{3}{4})}^{5}$ .

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$ , we have

$\begin{array}{l} {(x + a)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a x^{n - 1} + \dots + (\begin{matrix} n \\ j \end{matrix}) a^{j} x^{n - j} + \dots + (\begin{matrix} n \\ n \end{matrix}) a^{n} \\ {(x + a)}^{n} = Σ_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - j} a^{j} \end{array}$

Here $x = \frac{1}{4}, a = \frac{3}{4}$ and $n = 5$ . Applying Binomial theorem,

$(\begin{matrix} 5 \\ 0 \end{matrix}) {(\frac{1}{4})}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(\frac{1}{4})}^{4} (\frac{3}{4}) + (\begin{matrix} 5 \\ 2 \end{matrix}) {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{4} + (\begin{matrix} 5 \\ 5 \end{matrix}) {(\frac{3}{4})}^{5} = {(\frac{1}{4} + \frac{3}{4})}^{5} = {(1)}^{5} = 1$

Chapter 12.5, Problem 48AYU

Chapter 12.5, Problem 50AYU

Chapter 12 Solutions

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( 5 0 ) ( 1 4 ) 5 + ( 5 1 ) ( 1 4 ) 4 ( 3 4 ) + ( 5 2 ) ( 1 4 ) 3 ( 3 4 ) 2 + ( 5 3 ) ( 1 4 ) 2 ( 3 4 ) 3 + ( 5 4 ) ( 1 4 ) ( 3 4 ) 4 + ( 5 5 ) ( 3 4 ) 5 = ?

Solution Summary: The author explains the Binomial Theorem: Let x and a be real numbers.

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Answer to Problem 49AYU

Explanation of Solution

Chapter 12 Solutions

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