Explanation Given: Expression is given as ( n 0 ) + ( n 1 ) + ⋯ + ( n n ) = 2 n if n is a positive integer. It can be written as 2 n = ( 1 + 1 ) n 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 , a = 1 and n = n . Applying Binomial theore, ( 1 + 1 ) n = ( n 0 ) ( 1 ) n + ( n ...

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Chapter 12.5, Problem 47AYU

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To show: To show that $(\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + \dots + (\begin{matrix} n \\ n \end{matrix}) = 2^{n}$ if $n$ is a positive integer.

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Chapter 12.5, Problem 46AYU

Chapter 12.5, Problem 48AYU

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To show that ( n 0 ) + ( n 1 ) + ⋯ + ( n n ) = 2 n if n is a positive integer.

Solution Summary: The author explains that the numerical value of ( 1.001 ) 5 using the Binomial Theorem is 1.00501.

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To show: To show that $(\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + \dots + (\begin{matrix} n \\ n \end{matrix}) = 2^{n}$ if $n$ is a positive integer.

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Chapter 12 Solutions

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To show that ( n 0 ) + ( n 1 ) + ⋯ + ( n n ) = 2 n if n is a positive integer.

Solution Summary: The author explains that the numerical value of ( 1.001 ) 5 using the Binomial Theorem is 1.00501.

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To show: To show that (n0)+(n1)+⋯+(nn)=2n if n is a positive integer.

Want to see the full answer?

Chapter 12 Solutions

Additional Math Textbook Solutions

To show: To show that $(\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + \dots + (\begin{matrix} n \\ n \end{matrix}) = 2^{n}$ if $n$ is a positive integer.