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

Given information: We are given the first term of the sequence, a_(1)=3, and the recursive formula for the sequence, a_(n)=(a_(n-1))^2+n. To find a_(4), we need to find the values of a_(2), a_(3), and then a_(4) using the recursive formula.Find a_(2): First, let's find a_(2) using the recursive formula with n=2: a_(2) = (a_(1))^2 + 2 = (3)^2 + 2 = 9 + 2 = 11 So, a_(2) = 11.Find a_(3): Next, we find a_(3) using the recursive formula with n=3: a_(3) = (a_(2))^2 + 3 = (11)^2 + 3 = 121 + 3 = 124 So, a_(3) = 124.Find a_(4): Finally, we find a_(4) using the recursive formula with n=4: a_(4) = (a_(3))^2 + 4 = (124)^2 + 4 = 15376 + 4 = 15380 So, a_(4) = 15380.

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

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

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Precalculus

Find the roots of factored polynomials

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Q. If

a_{1}=3

and

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

then find the value of

a_{4}

\newline

Answer:

Given information: We are given the first term of the sequence, $a_{1}=3$ , and the recursive formula for the sequence, $a_{n}=(a_{n-1})^2+n$ . To find $a_{4}$ , we need to find the values of $a_{2}$ , $a_{3}$ , and then $a_{4}$ using the recursive formula.
Find $a_{2}$ : First, let's find $a_{2}$ using the recursive formula with $n=2$ : $a_{2} = (a_{1})^2 + 2$ $= (3)^2 + 2$ $= 9 + 2$ $= 11$ So, $a_{2} = 11$ .
Find $a_{3}$ : Next, we find $a_{3}$ using the recursive formula with $n=3$ : $a_{3} = (a_{2})^2 + 3$ $= (11)^2 + 3$ $= 121 + 3$ $= 124$ So, $a_{3} = 124$ .
Find $a_{4}$ : Finally, we find $a_{4}$ using the recursive formula with $n=4$ : $\newline$ $a_{4} = (a_{3})^2 + 4$ $\newline$ $= (124)^2 + 4$ $\newline$ $= 15376 + 4$ $\newline$ $= 15380$ $\newline$ So, $a_{4} = 15380$ .