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,

2, 5, 8, 11,

2, 3, 5, 11, 29,

2,6,18,54,

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

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

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 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 (ie, the first term is 2)

The recursive definition is 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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(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)