Which recursive sequence would produce the sequence 1,-2,-11,dots ? a_(1)=1 and a_(n)=2a_(n-1)-4 a_(1)=1 and a_(n)=3a_(n-1)-5 a_(1)=1 and a_(n)=-5a_(n-1)+3 a_(1)=1 and a_(n)=-4a_(n-1)+2

Test Recursive Formula: Let's start by testing each given recursive formula with the initial value a_(1)=1 to see which one produces the sequence 1, -2, -11, ... . We will begin with the first option: a_(1)=1 and a_(n)=2a_(n-1)-4 We need to calculate a_(2) using the initial value a_(1)=1. a_(2)=2a_(1)-4 a_(2)=2(1)-4 a_(2)=2-4 a_(2)=-2 So far, the sequence matches: 1, -2. Now let's calculate a_(3). a_(3)=2a_(2)-4 a_(3)=2(-2)-4 a_(3)=-4-4 a_(3)=-8 The third term we calculated is -8, but the third term in the given sequence is -11. Therefore, this recursive formula does not produce the given sequence.

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Which recursive sequence would produce the sequence
1,-2,-11,dots ?

a_(1)=1 and
a_(n)=2a_(n-1)-4

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

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

a_(1)=1 and
a_(n)=-4a_(n-1)+2

Which recursive sequence would produce the sequence $1,-2,-11, \ldots$ ? $\newline$ $a_{1}=1$ and $a_{n}=2 a_{n-1}-4$ $\newline$ $a_{1}=1$ and $a_{n}=3 a_{n-1}-5$ $\newline$ $a_{1}=1$ and $a_{n}=-5 a_{n-1}+3$ $\newline$ $a_{1}=1$ and $a_{n}=-4 a_{n-1}+2$

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Find derivatives of using multiple formulae

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Q. Which recursive sequence would produce the sequence

1,-2,-11, \ldots

\newline

a_{1}=1

and

a_{n}=2 a_{n-1}-4

\newline

a_{1}=1

and

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

\newline

a_{1}=1

and

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

\newline

a_{1}=1

and

a_{n}=-4 a_{n-1}+2

Test Recursive Formula: Let's start by testing each given recursive formula with the initial value $a_{1}=1$ to see which one produces the sequence $1, -2, -11, \ldots$ . $\newline$ We will begin with the first option: $\newline$ $a_{1}=1$ and $\newline$ $a_{n}=2a_{n-1}-4$ $\newline$ We need to calculate $a_{2}$ using the initial value $a_{1}=1$ . $\newline$ $a_{2}=2a_{1}-4$ $\newline$ $a_{2}=2(1)-4$ $\newline$ $a_{2}=2-4$ $\newline$ $a_{2}=-2$ $\newline$ So far, the sequence matches: $1, -2, -11, \ldots$ $0$ . Now let's calculate $1, -2, -11, \ldots$ $1$ . $\newline$ $1, -2, -11, \ldots$ $2$ $\newline$ $1, -2, -11, \ldots$ $3$ $\newline$ $1, -2, -11, \ldots$ $4$ $\newline$ $1, -2, -11, \ldots$ $5$ $\newline$ The third term we calculated is $1, -2, -11, \ldots$ $6$ , but the third term in the given sequence is $1, -2, -11, \ldots$ $7$ . Therefore, this recursive formula does not produce the given sequence.