Chapter 12.5, Problem 47AYU

If $n$ is a positive integer, show that, $\left(\begin{array}{l}n\\ 0\end{array}\right)+\left(\begin{array}{l}n\\ 1\end{array}\right)+\dots +\left(\begin{array}{l}n\\ n\end{array}\right)=2^n$

[Hint: $2^n={\left(1+1\right)}^n$; now use the Binomial Theorem.]

To determine

To show: To show that $\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)=2^n$ if $n$ is a positive integer.

Answer to Problem 47AYU

Solution:

The numerical value of ${(1.001)}^5$ using Binomial Theorem is $1.00501$

Explanation of Solution

Given:

Expression is given as $\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)=2^n$ 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

$\begin{array}{l}{\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\cdots +\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^n\\ {\left(x+a\right)}^n=\underset{k=0}{\overset{n}{{\displaystyle \Sigma }}}\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-j}{a}^j\end{array}$

Here $x=1,a=1$ and $n=n$. Applying Binomial theore,

${\left(1+1\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right){\left(1\right)}^n+\left(\begin{array}{c}n\\ 1\end{array}\right)\left(1\right){\left(1\right)}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right){\left(1\right)}^2{\left(1\right)}^{n-2}+\left(\begin{array}{c}n\\ 3\end{array}\right){\left(1\right)}^3{\left(1\right)}^{n-3}+\cdots +\left(\begin{array}{c}n\\ n-1 \end{array}\right){\left(1\right)}^{n-1}{\left(1\right)}^1+\left(\begin{array}{c}n\\ n\end{array}\right){\left(1\right)}^n$

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