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Posted 3 months ago
Consider the geometric sequence:
8,4,2,1,8,4,2,1, \ldots
If nn is an integer, which of these functions generate the sequence?
Choose all answers that apply:
A a(n)=8(12)na(n)=8\left(\frac{1}{2}\right)^{n} for n1n \geq 1
B b(n)=32(12)nb(n)=32\left(\frac{1}{2}\right)^{n} for n2n \geq 2
c. c(n)=64(12)nc(n)=64\left(\frac{1}{2}\right)^{n} for n3n \geq 3

D d(n)=128(12)nd(n)=128\left(\frac{1}{2}\right)^{n} for n4n \geq 4
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 2
For option A, a(n)=8(12)na(n)=8\left(\frac{1}{2}\right)^{n} for n1n \geq 1, we calculate the first four terms: a(1)=8(12)1=4a(1)=8\left(\frac{1}{2}\right)^{1}=4, a(2)=8(12)2=2a(2)=8\left(\frac{1}{2}\right)^{2}=2, a(3)=8(12)3=1a(3)=8\left(\frac{1}{2}\right)^{3}=1, and a(4)=8(12)4=12a(4)=8\left(\frac{1}{2}\right)^{4}=\frac{1}{2}. This does not match the sequence starting from n=1n=1
step 3
For option B, b(n)=32(12)nb(n)=32\left(\frac{1}{2}\right)^{n} for n2n \geq 2, we calculate the first four terms starting from n=2n=2: b(2)=32(12)2=8b(2)=32\left(\frac{1}{2}\right)^{2}=8, b(3)=32(12)3=4b(3)=32\left(\frac{1}{2}\right)^{3}=4, b(4)=32(12)4=2b(4)=32\left(\frac{1}{2}\right)^{4}=2, and b(5)=32(12)5=1b(5)=32\left(\frac{1}{2}\right)^{5}=1. This matches the sequence starting from n=2n=2
step 4
For option C, c(n)=64(12)nc(n)=64\left(\frac{1}{2}\right)^{n} for n3n \geq 3, we calculate the first four terms starting from n=3n=3: c(3)=64(12)3=8c(3)=64\left(\frac{1}{2}\right)^{3}=8, c(4)=64(12)4=4c(4)=64\left(\frac{1}{2}\right)^{4}=4, c(5)=64(12)5=2c(5)=64\left(\frac{1}{2}\right)^{5}=2, and c(6)=64(12)6=1c(6)=64\left(\frac{1}{2}\right)^{6}=1. This matches the sequence starting from n=3n=3
step 5
For option D, d(n)=128(12)nd(n)=128\left(\frac{1}{2}\right)^{n} for n4n \geq 4, we calculate the first four terms starting from n=4n=4: d(4)=128(12)4=8d(4)=128\left(\frac{1}{2}\right)^{4}=8, d(5)=128(12)5=4d(5)=128\left(\frac{1}{2}\right)^{5}=4, d(6)=128(12)6=2d(6)=128\left(\frac{1}{2}\right)^{6}=2, and d(7)=128(12)7=1d(7)=128\left(\frac{1}{2}\right)^{7}=1. This matches the sequence starting from n=4n=4
B, C, D
Key Concept
Geometric Sequence and Exponential Functions
The correct functions must generate the terms of the geometric sequence when evaluated at the specified values of nn. Options B, C, and D generate the sequence 8,4,2,1,8, 4, 2, 1, \ldots when starting at their respective values of nn.

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