If a_(1)=9 and a_(n)=4a_(n-1)-1 then find the value of a_(3). Answer:

Find a_(2): To find the value of a_(3), we need to first find the value of a_(2) using the recursive formula a_(n)=4a_(n-1)-1, with n=2.Calculate a_(2): Substitute n=2 into the recursive formula to get a_(2)=4a_(2-1)-1, which simplifies to a_(2)=4a_(1)-1.Find a_(3): Since we know that a_(1)=9, we can substitute this value into the expression for a_(2) to get a_(2)=4*9-1.Calculate a_(3): Calculate the value of a_(2) by performing the operations: a_(2)=4*9-1 = 36-1 = 35.Now that we have the value of a_(2), we can use it to find a_(3) using the same recursive formula a_(n)=4a_(n-1)-1, with n=3.Substitute n=3 into the recursive formula to get a_(3)=4a_(3-1)-1, which simplifies to a_(3)=4a_(2)-1.Since we have found that a_(2)=35, we can substitute this value into the expression for a_(3) to get a_(3)=4*35-1.Calculate the value of a_(3) by performing the operations: a_(3)=4*35-1 = 140-1 = 139.

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If
a_(1)=9 and
a_(n)=4a_(n-1)-1 then find the value of
a_(3).
Answer:

If $a_{1}=9$ and $a_{n}=4 a_{n-1}-1$ then find the value of $a_{3}$ . $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

Full solution

Q. If

a_{1}=9

and

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

then find the value of

a_{3}

\newline

Answer:

Find $a_{2}$ : To find the value of $a_{3}$ , we need to first find the value of $a_{2}$ using the recursive formula $a_{n}=4a_{n-1}-1$ , with $n=2$ .
Calculate $a_{2}$ : Substitute $n=2$ into the recursive formula to get $a_{2}=4a_{2-1}-1$ , which simplifies to $a_{2}=4a_{1}-1$ .
Find $a_{3}$ : Since we know that $a_{1}=9$ , we can substitute this value into the expression for $a_{2}$ to get $a_{2}=4\times9-1$ .
Calculate $a_{3}$ : Calculate the value of $a_{2}$ by performing the operations: $a_{2}=4\times 9-1 = 36-1 = 35$ .
Calculate $a_{3}$ : Calculate the value of $a_{2}$ by performing the operations: $a_{2}=4\times 9-1 = 36-1 = 35$ .Now that we have the value of $a_{2}$ , we can use it to find $a_{3}$ using the same recursive formula $a_{n}=4a_{n-1}-1$ , with $n=3$ .
Calculate $a_{3}$ : Calculate the value of $a_{2}$ by performing the operations: $a_{2}=4\times9-1 = 36-1 = 35$ .Now that we have the value of $a_{2}$ , we can use it to find $a_{3}$ using the same recursive formula $a_{n}=4a_{n-1}-1$ , with $n=3$ .Substitute $n=3$ into the recursive formula to get $a_{3}=4a_{3-1}-1$ , which simplifies to $a_{3}=4a_{2}-1$ .
Calculate $a_{3}$ : Calculate the value of $a_{2}$ by performing the operations: $a_{2}=4\times 9-1 = 36-1 = 35$ .Now that we have the value of $a_{2}$ , we can use it to find $a_{3}$ using the same recursive formula $a_{n}=4a_{n-1}-1$ , with $n=3$ .Substitute $n=3$ into the recursive formula to get $a_{3}=4a_{3-1}-1$ , which simplifies to $a_{3}=4a_{2}-1$ .Since we have found that $a_{2}$ $0$ , we can substitute this value into the expression for $a_{3}$ to get $a_{2}$ $2$ .
Calculate $a_{3}$ : Calculate the value of $a_{2}$ by performing the operations: $a_{2}=4\times 9-1 = 36-1 = 35$ .Now that we have the value of $a_{2}$ , we can use it to find $a_{3}$ using the same recursive formula $a_{n}=4a_{n-1}-1$ , with $n=3$ .Substitute $n=3$ into the recursive formula to get $a_{3}=4a_{3-1}-1$ , which simplifies to $a_{3}=4a_{2}-1$ .Since we have found that $a_{2}$ $0$ , we can substitute this value into the expression for $a_{3}$ to get $a_{2}$ $2$ .Calculate the value of $a_{3}$ by performing the operations: $a_{2}$ $4$ .