If a_(1)=2,a_(2)=0 and a_(n)=3a_(n-1)+3a_(n-2) then find the value of a_(4). Answer:

Calculate a_(3): To find the value of a_(4), we need to use the recursive formula a_(n)=3a_(n-1)+3a_(n-2) and the given initial conditions a_(1)=2 and a_(2)=0. We will start by finding a_(3).Find a_(4): Using the recursive formula, we calculate a_(3) as follows: a_(3) = 3a_(2) + 3a_(1) a_(3) = 3(0) + 3(2) a_(3) = 0 + 6 a_(3) = 6Now that we have a_(3), we can use it along with a_(2) to find a_(4) using the same recursive formula: a_(4) = 3a_(3) + 3a_(2) a_(4) = 3(6) + 3(0) a_(4) = 18 + 0 a_(4) = 18

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Precalculus

Find the roots of factored polynomials

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Q. If

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

and

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

then find the value of

a_{4}

\newline

Answer:

Calculate $a_{3}$ : To find the value of $a_{4}$ , we need to use the recursive formula $a_{n}=3a_{n-1}+3a_{n-2}$ and the given initial conditions $a_{1}=2$ and $a_{2}=0$ . We will start by finding $a_{3}$ .
Find $a_{4}$ : Using the recursive formula, we calculate $a_{3}$ as follows: $\newline$ $a_{3} = 3a_{2} + 3a_{1}$ $\newline$ $a_{3} = 3(0) + 3(2)$ $\newline$ $a_{3} = 0 + 6$ $\newline$ $a_{3} = 6$
Find $a_{4}$ : Using the recursive formula, we calculate $a_{3}$ as follows: $\newline$ $a_{3} = 3a_{2} + 3a_{1}$ $\newline$ $a_{3} = 3(0) + 3(2)$ $\newline$ $a_{3} = 0 + 6$ $\newline$ $a_{3} = 6$ Now that we have $a_{3}$ , we can use it along with $a_{2}$ to find $a_{4}$ using the same recursive formula: $\newline$ $a_{4} = 3a_{3} + 3a_{2}$ $\newline$ $a_{3}$ $0$ $\newline$ $a_{3}$ $1$ $\newline$ $a_{3}$ $2$