Question

PLEASE HELP!!!!!!!!!!!!!!!

1) Which sequences could be described by the recursive definition LaTeX: a_{n+1}=3\cdot a_n-1a n + 1 = 3 ⋅ a n − 1
Group of answer choices


2, 5, 14, 41, 122...

2, 5, 8, 11, 14...

2, 3, 5, 11, 29, 86...

2,6,18,54,162...





2) Given that the first term is LaTeX: t_1=2t 1 = 2, and the recursive definition is LaTeX: t_{n+1}=3^{\left(t_n\right)}t n + 1 = 3 ( t n ), what would be the 2nd term (LaTeX: t_2t 2).
Group of answer choices

3

9

27

81

Answer 1

an=3+an-1

2, 5, 14, 41, 122..

9

Explanation:

Answer 2

Problem 1)

The notation you wrote down is a bit odd. There seems to be a typo. Please double check.

Assuming you meant to write

`a_(n) = 3 a_(n-1)`

then the answer is choice D) 2, 6, 18, 54, 162, ...

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Why is this? Because each term is found by multiplying the previous term by 3

first term = 2

second term = (first term)*3 = 2*3 = 6

third term = (second term)*3 = 6*3 = 18

fourth term = (third term)*3 = 18*3 = 54

fifth term = (fourth term)*3 = 54*3 = 162

this pattern continues forever

Extra information: The closed form equation is
`a_(n) = 2(3)^(n-1)` where 2 is the first term, 3 is the common ratio, and n is a positive integer.

some more info: notice picking any term in sequence {2,6,18,54,162,...} and dividing it over its previous term leads to 3

6/2 = 3

18/6 = 3

54/18 = 3

162/54 = 3

None of the other sequences listed have this property of the adjacent terms dividing to a constant ratio. So none of the other sequences are geometric.

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Problem 2)

The first term is
`t_(1) = 2` (ie, the first term is 2)

The recursive definition is
`t_(n+1) = 3^{t_(n)}` basically telling us "to get the next term, raise the prior term as an exponent with 3 as the base"

The general template looking like this: nth term = 3^(term just before the nth term)

The first term is 2, so the next term must be 3^2 = 9

The third term would be 3^9 = 19683 and so on.

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Final Answer: choice B) 9