Use the series below to answer the following questions.
[tex]$\sum_{k=1}^6 4(5)^{k-1}$[/tex]
[tex]$a_1=$[/tex] $\square$ [tex]$r=$[/tex] $\square$ [tex]$n=$[/tex] $\square$
Answer (1)
The given series is ∑ k = 1 6 4 ( 5 ) k − 1 . This is a geometric series, and we can identify its parameters as follows:
First term a 1 : The series starts at k = 1 , so to find a 1 , substitute k = 1 into the expression: a 1 = 4 ( 5 ) 1 − 1 = 4 ( 5 ) 0 = 4 So, a 1 = 4 .
Common ratio r : The common ratio r is the factor by which each term is multiplied to get the next term. In this series, r = 5 As seen from the formula 4 ( 5 ) k − 1 , each term differs from the previous one by a factor of 5.
Number of terms n : The series goes from k = 1 to k = 6 , thus it contains 6 terms: n = 6
In summary, the parameters of the geometric series are:
a 1 = 4
r = 5
n = 6
This means that you have an initial term of 4, and each subsequent term is obtained by multiplying the previous term by 5, with a total of 6 terms in the series.
