Question
Consider the geometric sequence:
\[8, 4, 2, 1, \dots\]
If
\[n\] is an integer, which of these functions generate the sequence?
Choose all answers that apply:
Choose all answers that apply:
(Choice A)
\[a(n)=8\left(\dfrac{1}2\right)^n\] for
\[n\geq1\]
A
\[a(n)=8\left(\dfrac{1}2\right)^n\] for
\[n\geq1\]
(Choice B)
\[b(n)=32\left(\dfrac{1}2\right)^n\] for
\[n\geq2\]
B
\[b(n)=32\left(\dfrac{1}2\right)^n\] for
\[n\geq2\]
(Choice C)
\[c(n)=64\left(\dfrac{1}2\right)^n\] for
\[n\geq3\]
C
\[c(n)=64\left(\dfrac{1}2\right)^n\] for
\[n\geq3\]
(Choice D)
\[d(n)=128\left(\dfrac{1}2\right)^n\] for
\[n\geq4\]
D
\[d(n)=128\left(\dfrac{1}2\right)^n\] for
\[n\geq4\]
Related content
\[8, 4, 2, 1, \dots\]
If
\[n\] is an integer, which of these functions generate the sequence?
Choose all answers that apply:
Choose all answers that apply:
(Choice A)
\[a(n)=8\left(\dfrac{1}2\right)^n\] for
\[n\geq1\]
A
\[a(n)=8\left(\dfrac{1}2\right)^n\] for
\[n\geq1\]
(Choice B)
\[b(n)=32\left(\dfrac{1}2\right)^n\] for
\[n\geq2\]
B
\[b(n)=32\left(\dfrac{1}2\right)^n\] for
\[n\geq2\]
(Choice C)
\[c(n)=64\left(\dfrac{1}2\right)^n\] for
\[n\geq3\]
C
\[c(n)=64\left(\dfrac{1}2\right)^n\] for
\[n\geq3\]
(Choice D)
\[d(n)=128\left(\dfrac{1}2\right)^n\] for
\[n\geq4\]
D
\[d(n)=128\left(\dfrac{1}2\right)^n\] for
\[n\geq4\]
Related content
Answers
GPT-4o mini
The given geometric sequence is:
\[ 8, 4, 2, 1, \dots \]
This sequence can be represented with the first term \( a = 8 \) and a common ratio \( r = \frac{1}{2} \). This means the \( n \)-th term of the sequence can be expressed as:
\[
a_n = 8 \left(\frac{1}{2}\right)^{n-1}
\]
for \( n \geq 1 \). The terms of the sequence are:
- For \( n = 1 \): \( a_1 = 8 \left(\frac{1}{2}\right)^{1-1} = 8 \)
- For \( n = 2 \): \( a_2 = 8 \left(\frac{1}{2}\right)^{2-1} = 4 \)
- For \( n = 3 \): \( a_3 = 8 \left(\frac{1}{2}\right)^{3-1} = 2 \)
- For \( n = 4 \): \( a_4 = 8 \left(\frac{1}{2}\right)^{4-1} = 1 \)
Now we will evaluate each of the provided choices:
**Choice A**:
\[ a(n) = 8\left(\frac{1}{2}\right)^n \text{ for } n \geq 1 \]
This choice does **not** match the sequence at \( n = 1 \). It calculates:
- For \( n = 1 \): \( a(1) = 8\left(\frac{1}{2}\right)^1 = 4 \) (not 8)
So, **Choice A is incorrect.**
**Choice B**:
\[ b(n) = 32\left(\frac{1}{2}\right)^n \text{ for } n \geq 2 \]
Calculating this:
- For \( n = 2 \): \( b(2) = 32\left(\frac{1}{2}\right)^2 = 32 \times \frac{1}{4} = 8 \)
- For \( n = 3 \): \( b(3) = 32\left(\frac{1}{2}\right)^3 = 32 \times \frac{1}{8} = 4 \)
- For \( n = 4 \): \( b(4) = 32\left(\frac{1}{2}\right)^4 = 32 \times \frac{1}{16} = 2 \)
- For \( n = 5 \): \( b(5) = 32\left(\frac{1}{2}\right)^5 = 32 \times \frac{1}{32} = 1 \)
So, **Choice B is correct.**
**Choice C**:
\[ c(n) = 64\left(\frac{1}{2}\right)^n \text{ for } n \geq 3 \]
Calculating this:
- For \( n = 3 \): \( c(3) = 64\left(\frac{1}{2}\right)^3 = 64 \times \frac{1}{8} = 8 \) (not 2 as required)
Thus, **Choice C is incorrect.**
**Choice D**:
\[ d(n) = 128\left(\frac{1}{2}\right)^n \text{ for } n \geq 4 \]
Calculating this:
- For \( n = 4 \): \( d(4) = 128\left(\frac{1}{2}\right)^4 = 128 \times \frac{1}{16} = 8 \) (not 1 as required)
Thus, **Choice D is incorrect.**
In summary, the only choice that correctly generates the sequence is:
- **Choice B**: \( b(n) = 32\left(\frac{1}{2}\right)^n \text{ for } n \geq 2 \)
\[ 8, 4, 2, 1, \dots \]
This sequence can be represented with the first term \( a = 8 \) and a common ratio \( r = \frac{1}{2} \). This means the \( n \)-th term of the sequence can be expressed as:
\[
a_n = 8 \left(\frac{1}{2}\right)^{n-1}
\]
for \( n \geq 1 \). The terms of the sequence are:
- For \( n = 1 \): \( a_1 = 8 \left(\frac{1}{2}\right)^{1-1} = 8 \)
- For \( n = 2 \): \( a_2 = 8 \left(\frac{1}{2}\right)^{2-1} = 4 \)
- For \( n = 3 \): \( a_3 = 8 \left(\frac{1}{2}\right)^{3-1} = 2 \)
- For \( n = 4 \): \( a_4 = 8 \left(\frac{1}{2}\right)^{4-1} = 1 \)
Now we will evaluate each of the provided choices:
**Choice A**:
\[ a(n) = 8\left(\frac{1}{2}\right)^n \text{ for } n \geq 1 \]
This choice does **not** match the sequence at \( n = 1 \). It calculates:
- For \( n = 1 \): \( a(1) = 8\left(\frac{1}{2}\right)^1 = 4 \) (not 8)
So, **Choice A is incorrect.**
**Choice B**:
\[ b(n) = 32\left(\frac{1}{2}\right)^n \text{ for } n \geq 2 \]
Calculating this:
- For \( n = 2 \): \( b(2) = 32\left(\frac{1}{2}\right)^2 = 32 \times \frac{1}{4} = 8 \)
- For \( n = 3 \): \( b(3) = 32\left(\frac{1}{2}\right)^3 = 32 \times \frac{1}{8} = 4 \)
- For \( n = 4 \): \( b(4) = 32\left(\frac{1}{2}\right)^4 = 32 \times \frac{1}{16} = 2 \)
- For \( n = 5 \): \( b(5) = 32\left(\frac{1}{2}\right)^5 = 32 \times \frac{1}{32} = 1 \)
So, **Choice B is correct.**
**Choice C**:
\[ c(n) = 64\left(\frac{1}{2}\right)^n \text{ for } n \geq 3 \]
Calculating this:
- For \( n = 3 \): \( c(3) = 64\left(\frac{1}{2}\right)^3 = 64 \times \frac{1}{8} = 8 \) (not 2 as required)
Thus, **Choice C is incorrect.**
**Choice D**:
\[ d(n) = 128\left(\frac{1}{2}\right)^n \text{ for } n \geq 4 \]
Calculating this:
- For \( n = 4 \): \( d(4) = 128\left(\frac{1}{2}\right)^4 = 128 \times \frac{1}{16} = 8 \) (not 1 as required)
Thus, **Choice D is incorrect.**
In summary, the only choice that correctly generates the sequence is:
- **Choice B**: \( b(n) = 32\left(\frac{1}{2}\right)^n \text{ for } n \geq 2 \)