Chapter 8.3, Problem 67E

To determine

To write: the sum $2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}$ by using summation notation.

Answer to Problem 67E

The summation notation for the sum $2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}$ is $2{\displaystyle \sum _{n=1}^6{\left(-1\right)}^{n-1}{2}^{-2n}}$ .

Explanation of Solution

Given information :

Givensum is $2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}$

As the given series is $2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}$ .

The above series can be simplified as

  $\begin{array}{c}2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}=2\left(1-\frac{1}{4}+\frac{1}{16}-\mathrm{...}+\frac{1}{4096}\right)\\ =2\left(2^0-2^{-2}+2^{-4}+\mathrm{...}+2^{-12}\right)\\ =2{\displaystyle \sum _{n=1}^6{\left(-1\right)}^{n-1}{2}^{-2n}}\end{array}$

Hence,the summation notation of the sum $2-\frac{1}{2}+\frac{1}{8}-\mathrm{...}+\frac{1}{2048}$ is $2{\displaystyle \sum _{n=1}^6{\left(-1\right)}^{n-1}{2}^{-2n}}$ .