If a_(1)=9 and a_(n)=3a_(n-1)-3 then find the value of a_(5). Answer:

Find a_(2): Use the given recursive formula to find a_(2). The recursive formula is a_(n)=3a_(n-1)-3. We know that a_(1)=9, so we can find a_(2) by plugging n=2 into the formula. a_(2) = 3a_(2-1) - 3 = 3a_(1) - 3 = 3*9 - 3 = 27 - 3 = 24.Find a_(3): Use the recursive formula to find a_(3). Now that we have a_(2), we can find a_(3) by plugging n=3 into the formula. a_(3) = 3a_(3-1) - 3 = 3a_(2) - 3 = 3*24 - 3 = 72 - 3 = 69.Find a_(4): Use the recursive formula to find a_(4). With a_(3) found, we can find a_(4) by plugging n=4 into the formula. a_(4) = 3a_(4-1) - 3 = 3a_(3) - 3 = 3*69 - 3 = 207 - 3 = 204.Find a_(5): Use the recursive formula to find a_(5). Finally, we can find a_(5) by plugging n=5 into the formula. a_(5) = 3a_(5-1) - 3 = 3a_(4) - 3 = 3*204 - 3 = 612 - 3 = 609.

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Precalculus

Find the roots of factored polynomials

Full solution

Q. If

a_{1}=9

and

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

then find the value of

a_{5}

\newline

Answer:

Find $a_{2}$ : Use the given recursive formula to find $a_{2}$ . The recursive formula is $a_{n}=3a_{n-1}-3$ . We know that $a_{1}=9$ , so we can find $a_{2}$ by plugging $n=2$ into the formula. $a_{2} = 3a_{2-1} - 3 = 3a_{1} - 3 = 3\times9 - 3 = 27 - 3 = 24$ .
Find $a_{3}$ : Use the recursive formula to find $a_{3}$ . Now that we have $a_{2}$ , we can find $a_{3}$ by plugging $n=3$ into the formula. $a_{3} = 3a_{3-1} - 3 = 3a_{2} - 3 = 3\times24 - 3 = 72 - 3 = 69$ .
Find $a_{4}$ : Use the recursive formula to find $a_{4}$ . With $a_{3}$ found, we can find $a_{4}$ by plugging $n=4$ into the formula. $a_{4} = 3a_{4-1} - 3 = 3a_{3} - 3 = 3\times69 - 3 = 207 - 3 = 204$ .
Find $a_{5}$ : Use the recursive formula to find $a_{5}$ . Finally, we can find $a_{5}$ by plugging $n=5$ into the formula. $a_{5} = 3a_{5-1} - 3 = 3a_{4} - 3 = 3\times204 - 3 = 612 - 3 = 609$ .