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What are the values of [tex]\( r \)[/tex] and [tex]\( a_1 \)[/tex] for [tex]\( \sum_{k=1}^6 \frac{1}{4}(2)^{k-1} \)[/tex]?

A. [tex]\( r = \frac{1}{4}, \, a_1 = 6 \)[/tex]
B. [tex]\( r = \frac{1}{4}, \, a_1 = 2 \)[/tex]
C. [tex]\( r = 2, \, a_1 = \frac{1}{4} \)[/tex]
D. [tex]\( r = 2, \, a_1 = 6 \)[/tex]


Sagot :

To determine the values of [tex]\( r \)[/tex] and [tex]\( a_1 \)[/tex] for the given series [tex]\(\sum_{k=1}^6 \frac{1}{4}(2)^{k-1}\)[/tex], we need to identify the common ratio and the initial term in the series.

The series is given in the form:
[tex]\[ \sum_{k=1}^6 \frac{1}{4}(2)^{k-1} \][/tex]

Let's break it down step by step:

1. The series can be written in general form as:
[tex]\[ a_1 \cdot r^{k-1} \][/tex]

2. Here, the term where [tex]\( k = 1 \)[/tex] would be:
[tex]\[ \frac{1}{4} \cdot (2)^0 = \frac{1}{4} \][/tex]

So, the first term [tex]\( a_1 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].

3. The term [tex]\( k = 2 \)[/tex] would be:
[tex]\[ \frac{1}{4} \cdot (2)^1 = \frac{1}{4} \times 2 \][/tex]

The exponent in each term clearly indicates that the common ratio [tex]\( r \)[/tex] is 2. This is because each subsequent term is obtained by multiplying the previous term by 2.

Based on this, the values of [tex]\( r \)[/tex] and [tex]\( a_1 \)[/tex] are:
[tex]\[ r = 2 \][/tex]
[tex]\[ a_1 = \frac{1}{4} \][/tex]

So, the correct answer is:
[tex]\[ \boxed{r = 2; a_1 = \frac{1}{4}} \][/tex]
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