Chapter 8.1, Problem 9E

To determine

To find: Formula for the general term $a_n$ of the sequence.

Answer to Problem 9E

The formula for the general term of the sequence is $a_n={\left(-1\right)}^{n-1}\frac{2^{n-1}}{3^{n-1}}$ or $a_n={\left(-1\right)}^{n+1}\frac{2^{n-1}}{3^{n-1}}$ .

Explanation of Solution

Given information: The sequence is $\left\{1,-\frac{2}{3},\frac{4}{9},-\frac{8}{27},\mathrm{....}\right\}$

Calculation:

Here, $$

  $a_1=1,a_2=-\frac{2}{3},a_3=\frac{4}{9},a_4=-\frac{8}{27}$

The numerators are powers of 2, in first term numerator is 2⁰, in second term is 2¹, in third term is 2²;

So, in general the n^th term will have numerator 2ⁿ-1.

The denominators are powers of 3, in first term denominator is 3⁰, in second term denominator is 3¹, in third term denominator is 3²;

So, in general the n^th term will have denominator 3ⁿ-1.

The signs of the terms are alternately positive and negative, so the general term will be multiplied by ${\left(-1\right)}^{n-1}$ or ${\left(-1\right)}^{n+1}$ .

  $\therefore$ General term $a_n={\left(-1\right)}^{n-1}\frac{2^{n-1}}{3^{n-1}}$ or $a_n={\left(-1\right)}^{n+1}\frac{2^{n-1}}{3^{n-1}}$ .

Hence, the formula for the general term of the sequence is $a_n={\left(-1\right)}^{n-1}\frac{2^{n-1}}{3^{n-1}}$ or $a_n={\left(-1\right)}^{n+1}\frac{2^{n-1}}{3^{n-1}}$ .