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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]
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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