Question

prove that$\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+........+\left(\begin{array}{c}n\\ n\end{array}\right)=2^n$

Short answer

The Binomial theorem states that

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \end{gathered}$$

Step-by-step solution

Step 1. Given information

In this question $\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+........+\left(\begin{array}{c}n\\ n\end{array}\right)$is given

We have to prove that$\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+........+\left(\begin{array}{c}n\\ n\end{array}\right)=2^n$

Step 2. Description of proving R.H.S=L.H.S

Here by using the binomial theorem we can prove RHS =LHS , So we can rewrite$$\begin{gathered} 2^n={(1+1)}^{n}andbyapplyingbinomialtheoremonthisformweget, \\ {(1+1)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)1^n+\left(\begin{array}{c}n\\ 1\end{array}\right)\times 1\times 1^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)\times 1^2\times 1^{n-2}+\left(\begin{array}{c}n\\ 3\end{array}\right)\times 1^3\times 1^{n-3}+................+\left(\begin{array}{c}n\\ n\end{array}\right)\times 1^n\times 1^{n-n} \\ =\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+......+\left(\begin{array}{c}n\\ n\end{array}\right) \end{gathered}$$