Chapter 10.4, Problem 66HP

To determine

To write the below infinite series using sigma notation in two different ways:

  $3-6+12-\mathrm{.....}$

Answer to Problem 66HP

  ${\displaystyle {\sum }_{n=1}^{\infty }3\cdot {\left(-2\right)}^{n-1}}$ and ${\displaystyle {\sum }_{n=0}^{\infty }3\cdot {\left(-2\right)}^n}$ .

Explanation of Solution

Given:

  $3-6+12-\mathrm{.....}$

Concept Used:

An infinite geometric series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r can be written in sigma notation:

 1. $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}={\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}$

 2. $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}={\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}$

Calculation:

In order to write the below infinite series using sigma notation in two different ways:

  $3-6+12-\mathrm{.....}$

First find the first term and the common ratio of the series.

Now, from the given series it can be seen that it’s first term is 3 and the common ratio of the series, r , can be find by dividing the second term by the first term, as

  $r=\frac{a_2}{a_1}$

Since, here

  $a_2=-6,a_1=3$

Thus,

  $\begin{array}{c}r=\frac{-6}{3}\\ =-2\end{array}$

Thus, the given series can be written in the form $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ as

  $3-6+12-\mathrm{.....}=3+3{\left(-2\right)}^1+3{\left(-2\right)}^2+3{\left(-2\right)}^3+\mathrm{...}$

And the above series can be represented using sigma notation in two different ways as:

 1. $\begin{array}{c}3-6+12-\mathrm{.....}=3+3{\left(-2\right)}^1+3{\left(-2\right)}^2+3{\left(-2\right)}^3+\mathrm{...}\\ ={\displaystyle {\sum }_{n=1}^{\infty }3\cdot {\left(-2\right)}^{n-1}}\end{array}$

 2. $\begin{array}{c}3-6+12-\mathrm{.....}=3+3{\left(-2\right)}^1+3{\left(-2\right)}^2+3{\left(-2\right)}^3+\mathrm{...}\\ ={\displaystyle {\sum }_{n=0}^{\infty }3\cdot {\left(-2\right)}^n}\end{array}$

Thus, two different ways to write the given series using sigma notations are: ${\displaystyle {\sum }_{n=1}^{\infty }3\cdot {\left(-2\right)}^{n-1}}$ and ${\displaystyle {\sum }_{n=0}^{\infty }3\cdot {\left(-2\right)}^n}$ .