Which recursive sequence would produce the sequence 9,-32,91,dots ? a_(1)=9 and a_(n)=-3a_(n-1)-5 a_(1)=9 and a_(n)=-4a_(n-1)+4 a_(1)=9 and a_(n)=-5a_(n-1)-3 a_(1)=9 and a_(n)=4a_(n-1)-4

Identify First Term: Identify the first term of the sequence. The first term given for all options is a_(1) = 9.Calculate Second Term: Use the second option to calculate the second term. The second option is a_(n) = -4a_(n-1) + 4. Let's calculate a_(2) using a_(1) = 9: a_(2) = -4 * a_(1) + 4 = -4 * 9 + 4 = -36 + 4 = -32. This matches the second term of the sequence.Calculate Third Term: Use the second option to calculate the third term. Now, let's calculate a_(3) using a_(2) = -32: a_(3) = -4 * a_(2) + 4 = -4 * (-32) + 4 = 128 + 4 = 132. This does not match the third term of the sequence (91).

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Which recursive sequence would produce the sequence
9,-32,91,dots ?

a_(1)=9 and
a_(n)=-3a_(n-1)-5

a_(1)=9 and
a_(n)=-4a_(n-1)+4

a_(1)=9 and
a_(n)=-5a_(n-1)-3

a_(1)=9 and
a_(n)=4a_(n-1)-4

Which recursive sequence would produce the sequence $9,-32,91, \ldots$ ? $\newline$ $a_{1}=9$ and $a_{n}=-3 a_{n-1}-5$ $\newline$ $a_{1}=9$ and $a_{n}=-4 a_{n-1}+4$ $\newline$ $a_{1}=9$ and $a_{n}=-5 a_{n-1}-3$ $\newline$ $a_{1}=9$ and $a_{n}=4 a_{n-1}-4$

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Math Problems

Algebra 1

Identify a sequence as explicit or recursive

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Q. Which recursive sequence would produce the sequence

9,-32,91, \ldots

\newline

a_{1}=9

and

a_{n}=-3 a_{n-1}-5

\newline

a_{1}=9

and

a_{n}=-4 a_{n-1}+4

\newline

a_{1}=9

and

a_{n}=-5 a_{n-1}-3

\newline

a_{1}=9

and

a_{n}=4 a_{n-1}-4

Identify First Term: Identify the first term of the sequence. $\newline$ The first term given for all options is $a_{1} = 9$ .
Calculate Second Term: Use the second option to calculate the second term. $\newline$ The second option is $a_{n} = -4a_{n-1} + 4$ . Let's calculate $a_{2}$ using $a_{1} = 9$ : $\newline$ $a_{2} = -4 \times a_{1} + 4 = -4 \times 9 + 4 = -36 + 4 = -32$ . $\newline$ This matches the second term of the sequence.
Calculate Third Term: Use the second option to calculate the third term. $\newline$ Now, let's calculate $a_{3}$ using $a_{2} = -32$ : $\newline$ $a_{3} = -4 \times a_{2} + 4 = -4 \times (-32) + 4 = 128 + 4 = 132$ . $\newline$ This does not match the third term of the sequence ( $91$ ).