Q. 1.13 - A First Course in Probability

Question

Show that, for n > 0,

$\sum _{i=0}^n{(-1)}^i\left(\begin{array}{c}n\\ i\end{array}\right)=0$

Hint: Use the binomial theorem.

Step-by-Step Solution

Answer

It is proved that, for $n>0$

$\sum _{i=0}^n{(-1)}^i\left(\begin{array}{c}n\\ i\end{array}\right)=0$

Step 1. Given information.

Here, we have to prove that, for $n>0$

Step 2. State the Binomial Theorem

According to the Binomial Theorem

${(x+y)}^n=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)x^i{y}^{n-i}$

Step 3. Prove that ∑ i = 0 n ( - 1 ) i n i = 0

Let us assume $x=-1 \text{and} y=1$ in ${(x+y)}^n=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)x^i{y}^{n-i}$

$$\begin{gathered} {(-1+1)}^n=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^i{\left(1\right)}^{n-i} \\ \Rightarrow {\left(0\right)}^n=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^i\times 1 \\ \Rightarrow 0=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^i \end{gathered}$$

Therefore, it is proved that $\sum _{i=0}^n{(-1)}^i\left(\begin{array}{c}n\\ i\end{array}\right)=0$