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

Given terms and formula: We are given the first term of the sequence, a_(1) = 3, and the recursive formula for the sequence, a_(n+1) = (a_(n))^2 - 1. To find a_(3), we first need to find a_(2) using the given formula.Find a_(2): Using the recursive formula a_(n+1) = (a_(n))^2 - 1, we substitute n = 1 to find a_(2): a_(2) = (a_(1))^2 - 1 a_(2) = (3)^2 - 1 a_(2) = 9 - 1 a_(2) = 8Find a_(3): Now that we have a_(2), we can use it to find a_(3) using the same recursive formula: a_(3) = (a_(2))^2 - 1 a_(3) = (8)^2 - 1 a_(3) = 64 - 1 a_(3) = 63

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

If $a_{1}=3$ and $a_{n+1}=\left(a_{n}\right)^{2}-1$ then find the value of $a_{3}$ . $\newline$ Answer:

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Algebra 2

Transformations of functions

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Q. If

a_{1}=3

and

a_{n+1}=\left(a_{n}\right)^{2}-1

then find the value of

a_{3}

\newline

Answer:

Given terms and formula: We are given the first term of the sequence, $a_{1} = 3$ , and the recursive formula for the sequence, $a_{n+1} = (a_{n})^2 - 1$ . To find $a_{3}$ , we first need to find $a_{2}$ using the given formula.
Find $a_{2}$ : Using the recursive formula $a_{n+1} = (a_{n})^2 - 1$ , we substitute $n = 1$ to find $a_{2}$ :
$a_{2} = (a_{1})^2 - 1$
$a_{2} = (3)^2 - 1$
$a_{2} = 9 - 1$
$a_{2} = 8$
Find $a_{3}$ : Now that we have $a_{2}$ , we can use it to find $a_{3}$ using the same recursive formula: $\newline$ $a_{3} = (a_{2})^2 - 1$ $\newline$ $a_{3} = (8)^2 - 1$ $\newline$ $a_{3} = 64 - 1$ $\newline$ $a_{3} = 63$