Exercise 23 Page 575

Practice makes perfect

Let's start by considering the given geometric series.

We can see that the common ratio is 12. Before we find the sum, we need to be sure that the series converges. We check this by calculating its absolute value. |r|=| 1/2|=1/2 Because the absolute value of the common ratio is less than 1, we know that the series converges. To find its sum, we will substitute r= 12 and a_1= 6 into the formula for the sum of an infinite geometric series.

S=a_1/1-r

SubstituteII

r= 1/2, a_1= 6

S=6/1- 12

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Evaluate right-hand side

OneToFrac

Rewrite 1 as 2/2

S=6/22- 12

SubFrac

Subtract fractions

S=6/1/2

DivByFracD

a/b/c= a * c/b

S=12/1

DivByOne

a/1=a

S=12

Consider the given infinite geometric series written in summation notation.

∑^(∞)_(k=1) ( 1/3)^k Before we can find the sum of an infinite geometric series, we need to be sure that the series converges. We can check this by calculating the absolute value of its common ratio. For this series, we can see that the common ratio is 13. |r|=| 1/3|=1/3 Because the absolute value of the common ratio is less than 1, we know that the series converges. To calculate its sum, we first need to find its first term.

t(k)=(1/3)^k

Substitute

k= 1

t( 1)=(1/3)^1

ExponentOne

a^1=a

t(1)= 1/3

Now we are ready to find the sum! We will substitute r= 13 and t(1)= 13 into the formula for the sum of an infinite geometric series.

S=t(1)/1-r

SubstituteII

r= 1/3, t(1)= 1/3

S=13/1- 13