What is the nth term of the geometric sequence 4, 8, 16, 32, ... ?

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What is the nth term of the geometric sequence 4, 8, 16, 32, ... ?

Mathematics
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A. 2(4)^N-1 B. 4(2)^N-1 C. 2(N)^4 D. 4(N)^2
formula is (first term)*(common ratio)^(N-1) what is the first term?
\[a _{n}\]

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Other answers:

not quite, take a look at your sequence what is the first number in the list?
ditto
A. \[A. a_{n}=2(4)^{n-1} B. a_{n}=4(2)^{n-1} C.a_{n}=2(n)^{4} D. a_{n}=4(n)^{2}\]2(4)^N-1 B. 4(2)^N-1 C. 2(N)^4 D. 4(N)^2
would you mind answering my question, please? your sequence is 4, 8, 16, 32, ... , right? what is the first number in the sequence?
idk
which number comes first in the list 4, 8, 16, 32, ... ?
4
right! that means our first term is 4 our sequence is 4, 8, 16, 32, ... each consecutive number gets multiplied by 2 to get to the next number, right? our common ratio is 2 so, first term = 4, and common ratio = 2 now, let's look at the formula (first term)*(common ratio)^(N-1) can you tell me which one is the answer now? :)
A
not quite, check carefully (first term)*(common ratio)^(N-1) first term is 4 common ratio is 2
B
much better, good job
What is the tenth term of the geometric sequence that has a common ratio of 1/3 and 36 as its fifth term? A. 4/7 B. 27/4 C. 4/81 D. 1/36

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