If a_(1)=3 and a_(n+1)=-2a_(n)-1 then find the value of a_(4). Answer:

Initialize with a_(1): To find the value of a_(4), we need to use the recursive formula a_(n+1)=-2a_(n)-1 starting from n=1 where a_(1)=3.Find a_(2): First, let's find a_(2) using the formula with n=1: a_(2) = -2a_(1) - 1 = -2*3 - 1 = -6 - 1 = -7.Find a_(3): Next, we find a_(3) using the formula with n=2: a_(3) = -2a_(2) - 1 = -2*(-7) - 1 = 14 - 1 = 13.Find a_(4): Finally, we find a_(4) using the formula with n=3: a_(4) = -2a_(3) - 1 = -2*13 - 1 = -26 - 1 = -27.

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If
a_(1)=3 and
a_(n+1)=-2a_(n)-1 then find the value of
a_(4).
Answer:

If $a_{1}=3$ and $a_{n+1}=-2 a_{n}-1$ then find the value of $a_{4}$ . $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

Full solution

Q. If

a_{1}=3

and

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

then find the value of

a_{4}

\newline

Answer:

Initialize with $a_{1}$ : To find the value of $a_{4}$ , we need to use the recursive formula $a_{n+1}=-2a_{n}-1$ starting from $n=1$ where $a_{1}=3$ .
Find $a_{2}$ : First, let's find $a_{2}$ using the formula with $n=1$ : $a_{2} = -2a_{1} - 1 = -2\times3 - 1 = -6 - 1 = -7$ .
Find $a_{3}$ : Next, we find $a_{3}$ using the formula with $n=2$ : $a_{3} = -2a_{2} - 1 = -2*(-7) - 1 = 14 - 1 = 13$ .
Find $a_{4}$ : Finally, we find $a_{4}$ using the formula with $n=3$ : $a_{4} = -2a_{3} - 1 = -2\times13 - 1 = -26 - 1 = -27$ .