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Sagot :
Let's tackle each part of the question systematically.
### a) Write down the nth term in this series.
To determine the nth term of the given geometric series, let’s first identify the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]).
The series is [tex]\(16, 8, 4, \ldots\)[/tex]
- The first term ([tex]\(a\)[/tex]) is 16.
- The common ratio ([tex]\(r\)[/tex]) can be found by dividing the second term by the first term: [tex]\(r = \frac{8}{16} = \frac{1}{2}\)[/tex].
The nth term of a geometric series is given by:
[tex]\[ t_n = a \cdot (r^{n-1}) \][/tex]
Substitute [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]:
[tex]\[ t_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \][/tex]
### b) How many terms should be added to get a sum that exceeds 31?
The sum of the first [tex]\(n\)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For our series, [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]. We want to find the smallest [tex]\(n\)[/tex] such that the sum [tex]\(S_n\)[/tex] exceeds 31. This translates into:
[tex]\[ 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} > 31 \][/tex]
We simplify the denominator:
[tex]\[ 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = 32 \cdot (1 - \left(\frac{1}{2}\right)^n) \][/tex]
Now:
[tex]\[ 32 \cdot (1 - \left(\frac{1}{2}\right)^n) > 31 \][/tex]
[tex]\[ 1 - \left(\frac{1}{2}\right)^n > \frac{31}{32} \][/tex]
[tex]\[ \left(\frac{1}{2}\right)^n < \frac{1}{32} \][/tex]
Since [tex]\(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\)[/tex], we find that for [tex]\(n = 6\)[/tex]:
[tex]\[ \left(\frac{1}{2}\right)^6 = \frac{1}{64} \][/tex]
[tex]\[ 1 - \left(\frac{1}{2}\right)^6 > 1 - \frac{1}{32} \][/tex]
So, [tex]\(6\)[/tex] terms are needed to exceed a sum of 31.
### c) Calculate the sum to infinity, if it exists.
The sum to infinity of a geometric series exists if the common ratio [tex]\(r\)[/tex] satisfies [tex]\(|r| < 1\)[/tex].
Given [tex]\(r = \frac{1}{2}\)[/tex], which is indeed less than 1 in absolute value, the sum to infinity, [tex]\(S_\infty\)[/tex], is given by:
[tex]\[ S_\infty = \frac{a}{1 - r} \][/tex]
Substituting [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]:
[tex]\[ S_\infty = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 \][/tex]
### d) Write the series in Sigma notation.
The geometric series can be expressed in Sigma notation as:
[tex]\[ \sum_{n=1}^{\infty} 16 \left(\frac{1}{2}\right)^{n-1} \][/tex]
In conclusion:
- The nth term of the series is [tex]\( t_n = 16 \left(\frac{1}{2}\right)^{n-1} \)[/tex].
- It takes 6 terms for the sum to exceed 31.
- The sum to infinity is 32.
- The series in Sigma notation is [tex]\( \sum_{n=1}^{\infty} 16 \left(\frac{1}{2}\right)^{n-1} \)[/tex].
### a) Write down the nth term in this series.
To determine the nth term of the given geometric series, let’s first identify the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]).
The series is [tex]\(16, 8, 4, \ldots\)[/tex]
- The first term ([tex]\(a\)[/tex]) is 16.
- The common ratio ([tex]\(r\)[/tex]) can be found by dividing the second term by the first term: [tex]\(r = \frac{8}{16} = \frac{1}{2}\)[/tex].
The nth term of a geometric series is given by:
[tex]\[ t_n = a \cdot (r^{n-1}) \][/tex]
Substitute [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]:
[tex]\[ t_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \][/tex]
### b) How many terms should be added to get a sum that exceeds 31?
The sum of the first [tex]\(n\)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For our series, [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]. We want to find the smallest [tex]\(n\)[/tex] such that the sum [tex]\(S_n\)[/tex] exceeds 31. This translates into:
[tex]\[ 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} > 31 \][/tex]
We simplify the denominator:
[tex]\[ 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = 32 \cdot (1 - \left(\frac{1}{2}\right)^n) \][/tex]
Now:
[tex]\[ 32 \cdot (1 - \left(\frac{1}{2}\right)^n) > 31 \][/tex]
[tex]\[ 1 - \left(\frac{1}{2}\right)^n > \frac{31}{32} \][/tex]
[tex]\[ \left(\frac{1}{2}\right)^n < \frac{1}{32} \][/tex]
Since [tex]\(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\)[/tex], we find that for [tex]\(n = 6\)[/tex]:
[tex]\[ \left(\frac{1}{2}\right)^6 = \frac{1}{64} \][/tex]
[tex]\[ 1 - \left(\frac{1}{2}\right)^6 > 1 - \frac{1}{32} \][/tex]
So, [tex]\(6\)[/tex] terms are needed to exceed a sum of 31.
### c) Calculate the sum to infinity, if it exists.
The sum to infinity of a geometric series exists if the common ratio [tex]\(r\)[/tex] satisfies [tex]\(|r| < 1\)[/tex].
Given [tex]\(r = \frac{1}{2}\)[/tex], which is indeed less than 1 in absolute value, the sum to infinity, [tex]\(S_\infty\)[/tex], is given by:
[tex]\[ S_\infty = \frac{a}{1 - r} \][/tex]
Substituting [tex]\(a = 16\)[/tex] and [tex]\(r = \frac{1}{2}\)[/tex]:
[tex]\[ S_\infty = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 \][/tex]
### d) Write the series in Sigma notation.
The geometric series can be expressed in Sigma notation as:
[tex]\[ \sum_{n=1}^{\infty} 16 \left(\frac{1}{2}\right)^{n-1} \][/tex]
In conclusion:
- The nth term of the series is [tex]\( t_n = 16 \left(\frac{1}{2}\right)^{n-1} \)[/tex].
- It takes 6 terms for the sum to exceed 31.
- The sum to infinity is 32.
- The series in Sigma notation is [tex]\( \sum_{n=1}^{\infty} 16 \left(\frac{1}{2}\right)^{n-1} \)[/tex].
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