Chapter F, Problem 25E

To determine

To find: The value of the sum ${\displaystyle \sum _{n=1}^{20}{\left(-1\right)}^n}$.

Answer to Problem 25E

The value of the sum ${\displaystyle \sum _{n=1}^{20}{\left(-1\right)}^n}$ is $0$.

Explanation of Solution

Definition used:

If $a_m,a_{m+1},\mathrm{...},a_n$ are real numbers and m and n are integers such that $m\le n$, then ${\displaystyle \sum _{i=m}^n{a}_i=a_m+}{a}_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_n$.

Calculation:

By the above definition, the sum ${\displaystyle \sum _{n=1}^{20}{\left(-1\right)}^n}$ expressed as follows.

$\begin{array}{c}{\displaystyle \sum _{n=1}^{20}{\left(-1\right)}^n}=\left\{\begin{array}{c}{\left(-1\right)}^1+{\left(-1\right)}^2+{\left(-1\right)}^3+{\left(-1\right)}^4+{\left(-1\right)}^5+{\left(-1\right)}^6+{\left(-1\right)}^7+{\left(-1\right)}^8+{\left(-1\right)}^9\\ +{\left(-1\right)}^{10}++{\left(-1\right)}^{11}+{\left(-1\right)}^{12}+{\left(-1\right)}^{13}+{\left(-1\right)}^{14}+{\left(-1\right)}^{15}+{\left(-1\right)}^{16}+{\left(-1\right)}^{17}+\\ {\left(-1\right)}^{18}+{\left(-1\right)}^{19}+{\left(-1\right)}^{20}\end{array}\right\}\\ =-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1\\ =0\end{array}$

Thus value of the sum ${\displaystyle \sum _{n=1}^{20}{\left(-1\right)}^n}$ is $0$.