Asked by Jaz
Evaluate sigma-summation Underscript lower i equals 1 overscript infinity EndScripts 26 left parenthesis start fraction 1 over 5 end fraction right parenthesis superscript lower i baseline.
(1 point)
Responses
26
26
6.5
6.5
32.5
32.5
130
(1 point)
Responses
26
26
6.5
6.5
32.5
32.5
130
Answers
Answered by
GPT-4o mini
To evaluate the summation
\[
\sum_{i=1}^{\infty} 26 \left(\frac{1}{5}\right)^i,
\]
we recognize this as a geometric series of the form
\[
\sum_{i=0}^{\infty} ar^i,
\]
where \(a\) is the first term and \(r\) is the common ratio. In our case:
- The first term \(a\) for \(i = 1\) is \(26 \frac{1}{5} = \frac{26}{5}\),
- The common ratio \(r = \frac{1}{5}\).
The formula for the sum of an infinite geometric series is given by:
\[
S = \frac{a}{1 - r},
\]
provided that \(|r| < 1\). Here, \( |r| = \frac{1}{5} < 1 \), so we can use the formula:
Calculating the sum:
\[
S = \frac{\frac{26}{5}}{1 - \frac{1}{5}} = \frac{\frac{26}{5}}{\frac{4}{5}} = \frac{26}{5} \cdot \frac{5}{4} = \frac{26}{4} = 6.5.
\]
Thus, the value of the summation is
\[
\sum_{i=1}^{\infty} 26 \left(\frac{1}{5}\right)^i = 6.5.
\]
The correct response is:
**6.5**
\[
\sum_{i=1}^{\infty} 26 \left(\frac{1}{5}\right)^i,
\]
we recognize this as a geometric series of the form
\[
\sum_{i=0}^{\infty} ar^i,
\]
where \(a\) is the first term and \(r\) is the common ratio. In our case:
- The first term \(a\) for \(i = 1\) is \(26 \frac{1}{5} = \frac{26}{5}\),
- The common ratio \(r = \frac{1}{5}\).
The formula for the sum of an infinite geometric series is given by:
\[
S = \frac{a}{1 - r},
\]
provided that \(|r| < 1\). Here, \( |r| = \frac{1}{5} < 1 \), so we can use the formula:
Calculating the sum:
\[
S = \frac{\frac{26}{5}}{1 - \frac{1}{5}} = \frac{\frac{26}{5}}{\frac{4}{5}} = \frac{26}{5} \cdot \frac{5}{4} = \frac{26}{4} = 6.5.
\]
Thus, the value of the summation is
\[
\sum_{i=1}^{\infty} 26 \left(\frac{1}{5}\right)^i = 6.5.
\]
The correct response is:
**6.5**