If a_(1)=3,a_(2)=0 and a_(n)=3a_(n-1)+3a_(n-2) then find the value of a_(5). Answer:

Initialize Recursive Formula: To find a_(5), we need to use the recursive formula a_(n)=3a_(n-1)+3a_(n-2) to find the values of a_(3) and a_(4) first.Calculate a_(3): Using the given values, let's calculate a_(3): a_(3) = 3a_(2) + 3a_(1) = 3*0 + 3*3 = 0 + 9 = 9.Calculate a_(4): Now, let's calculate a_(4) using the values of a_(3) and a_(2): a_(4) = 3a_(3) + 3a_(2) = 3*9 + 3*0 = 27 + 0 = 27.Find a_(5): Finally, we can find a_(5) using the values of a_(4) and a_(3): a_(5) = 3a_(4) + 3a_(3) = 3*27 + 3*9 = 81 + 27 = 108.

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If
a_(1)=3,a_(2)=0 and
a_(n)=3a_(n-1)+3a_(n-2) then find the value of
a_(5).
Answer:

If $a_{1}=3, a_{2}=0$ and $a_{n}=3 a_{n-1}+3 a_{n-2}$ then find the value of $a_{5}$ . $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

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Q. If

a_{1}=3, a_{2}=0

and

a_{n}=3 a_{n-1}+3 a_{n-2}

then find the value of

a_{5}

\newline

Answer:

Initialize Recursive Formula: To find $a_{5}$ , we need to use the recursive formula $a_{n}=3a_{n-1}+3a_{n-2}$ to find the values of $a_{3}$ and $a_{4}$ first.
Calculate $a_{3}$ : Using the given values, let's calculate $a_{3}$ : $a_{3} = 3a_{2} + 3a_{1} = 3\times 0 + 3\times 3 = 0 + 9 = 9$ .
Calculate $a_{4}$ : Now, let's calculate $a_{4}$ using the values of $a_{3}$ and $a_{2}$ : $a_{4} = 3a_{3} + 3a_{2} = 3\times9 + 3\times0 = 27 + 0 = 27$ .
Find $a_{5}$ : Finally, we can find $a_{5}$ using the values of $a_{4}$ and $a_{3}$ : $a_{5} = 3a_{4} + 3a_{3} = 3\times27 + 3\times9 = 81 + 27 = 108$ .