Question

If n is a positive integer, show that

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

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 $\left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+........+{(-1)}^n\left(\begin{array}{c}n\\ n\end{array}\right)isgiven$

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

. Description of proving the  given question

By using the binomial theorem we can prove this given statement

According to the Binomial theorem

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+............+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ putx=1anda=-1thenbyapplyingtheorem,wecanexpand{(1-1)}^n{(1-1)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)1^n+\left(\begin{array}{c}n\\ 1\end{array}\right)(-1)1^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right){(-1)}^2{1}^{n-2}+............+\left(\begin{array}{c}n\\ n\end{array}\right){(-1)}^n{1}^{n-n}=\left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+............+{(-1)}^n\left(\begin{array}{c}n\\ n\end{array}\right)but{(1-1)}^n={\left(0\right)}^n\sin ce(1-1)=0 \\ =0 \\ hence\left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+............+{(-1)}^n\left(\begin{array}{c}n\\ n\end{array}\right)=0 \end{gathered}$$