Chapter 8.1, Problem 47E

To determine

To write an expression for the apparent $n$ th term of the sequence.

Answer to Problem 47E

The apparent $n$ th term of the sequence is $a_n=\frac{{\left(-1\right)}^{n+1}}{2^n}$

Explanation of Solution

Given information:

Given sequence

  $\frac{1}{2},\frac{-1}{4},\frac{1}{8},\frac{-1}{16},\mathrm{...}$

Calculation:

Consider the sequence $\frac{1}{2},\frac{-1}{4},\frac{1}{8},\frac{-1}{16},\mathrm{...}$

For $n=1,2,3,4,\mathrm{5...}n$

The terms are $\frac{1}{2},\frac{-1}{4},\frac{1}{8},\frac{-1}{16},\mathrm{...}{a}_n$

Rewrite the terms as shown below,

  $\begin{array}{c}\frac{1}{2},\frac{-1}{4},\frac{1}{8},\frac{-1}{16},\mathrm{...}{a}_n=\frac{{\left(-1\right)}^2}{2},\frac{{\left(-1\right)}^3}{4},\frac{{\left(-1\right)}^4}{8},\frac{{\left(-1\right)}^5}{16},\mathrm{...}{a}_n\\ =\frac{{\left(-1\right)}^{1+1}}{2^1},\frac{{\left(-1\right)}^{2+1}}{2^2},\frac{{\left(-1\right)}^{3+1}}{2^3},\frac{{\left(-1\right)}^{4+1}}{2^4},\mathrm{...}{a}_n\end{array}$

Thus pattern followed is such that each term at even position is negative. Each term is a reciprocal of 2 raise to the power on $n$ .

Therefore, the general term is $a_n=\frac{{\left(-1\right)}^{n+1}}{2^n}$