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

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

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Transformations of functions

Full solution

Q. If

a_{1}=3

and

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

then find the value of

a_{4}

.

\newline

Answer:

Identify First Term: Identify the first term in the sequence. $\newline$ The first term $a_{1}$ is given as $3$ .
Find Second Term: Use the recursive formula to find the second term $a_{2}$ . The recursive formula is $a_{n+1} = (a_{n})^2 - 1$ . Substitute $n=1$ into the formula to find $a_{2}$ . $a_{2} = (a_{1})^2 - 1 = (3)^2 - 1 = 9 - 1 = 8$ .
Find Third Term: Use the recursive formula to find the third term $a_{3}$ . Substitute $n=2$ into the formula to find $a_{3}$ . $a_{3} = (a_{2})^2 - 1 = (8)^2 - 1 = 64 - 1 = 63$ .
Find Fourth Term: Use the recursive formula to find the fourth term $a_{4}$ . Substitute $n=3$ into the formula to find $a_{4}$ . $a_{4} = (a_{3})^2 - 1 = (63)^2 - 1 = 3969 - 1 = 3968$ .

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