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

Find a_(2) Value: Determine the value of a_(2) using the given recursive formula. The recursive formula is a_(n)=(a_(n-1))^2+n. Since we know that a_(1)=4, we can find a_(2) by plugging n=2 into the formula: a_(2) = (a_(1))^2 + 2 a_(2) = (4)^2 + 2 a_(2) = 16 + 2 a_(2) = 18Find a_(3) Value: Determine the value of a_(3) using the value of a_(2) found in Step 1. Now that we have a_(2)=18, we can find a_(3) by plugging n=3 into the formula: a_(3) = (a_(2))^2 + 3 a_(3) = (18)^2 + 3 a_(3) = 324 + 3 a_(3) = 327

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

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

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Precalculus

Find the roots of factored polynomials

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Q. If

a_{1}=4

and

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

then find the value of

a_{3}

\newline

Answer:

Find $a_{2}$ Value: Determine the value of $a_{2}$ using the given recursive formula. $\newline$ The recursive formula is $a_{n}=(a_{n-1})^2+n$ . Since we know that $a_{1}=4$ , we can find $a_{2}$ by plugging $n=2$ into the formula: $\newline$ $a_{2} = (a_{1})^2 + 2$ $\newline$ $a_{2} = (4)^2 + 2$ $\newline$ $a_{2} = 16 + 2$ $\newline$ $a_{2} = 18$
Find $a_{3}$ Value: Determine the value of $a_{3}$ using the value of $a_{2}$ found in Step $1$ . $\newline$ Now that we have $a_{2}=18$ , we can find $a_{3}$ by plugging $n=3$ into the formula: $\newline$ $a_{3} = (a_{2})^2 + 3$ $\newline$ $a_{3} = (18)^2 + 3$ $\newline$ $a_{3} = 324 + 3$ $\newline$ $a_{3} = 327$