If a_(1)=5,a_(2)=2 and a_(n)=3a_(n-1)+2a_(n-2) then find the value of a_(4). Answer:

Understand Recursive Formula: Understand the recursive formula and initial conditions. The recursive formula given is a_(n)=3a_(n-1)+2a_(n-2), which means each term is generated by multiplying the previous term by 3 and the term before that by 2, then adding them together. We are given a_(1)=5 and a_(2)=2 as initial conditions.Find a_(3) Value: Find the value of a_(3) using the recursive formula. Using the formula a_(n)=3a_(n-1)+2a_(n-2), we substitute n=3 to find a_(3). a_(3) = 3a_(2) + 2a_(1) a_(3) = 3(2) + 2(5) a_(3) = 6 + 10 a_(3) = 16Find a_(4) Value: Find the value of a_(4) using the recursive formula and the values of a_(2) and a_(3). Now we use the formula a_(n)=3a_(n-1)+2a_(n-2) with n=4, a_(3)=16, and a_(2)=2. a_(4) = 3a_(3) + 2a_(2) a_(4) = 3(16) + 2(2) a_(4) = 48 + 4 a_(4) = 52

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If
a_(1)=5,a_(2)=2 and
a_(n)=3a_(n-1)+2a_(n-2) then find the value of
a_(4).
Answer:

If $a_{1}=5, a_{2}=2$ and $a_{n}=3 a_{n-1}+2 a_{n-2}$ then find the value of $a_{4}$ . $\newline$ Answer:

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Math Problems

Grade 8

Write and solve direct variation equations

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Q. If

a_{1}=5, a_{2}=2

and

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

then find the value of

a_{4}

\newline

Answer:

Understand Recursive Formula: Understand the recursive formula and initial conditions. $\newline$ The recursive formula given is $a_{n}=3a_{n-1}+2a_{n-2}$ , which means each term is generated by multiplying the previous term by $3$ and the term before that by $2$ , then adding them together. We are given $a_{1}=5$ and $a_{2}=2$ as initial conditions.
Find $a_{3}$ Value: Find the value of $a_{3}$ using the recursive formula. $\newline$ Using the formula $a_{n}=3a_{n-1}+2a_{n-2}$ , we substitute $n=3$ to find $a_{3}$ . $\newline$ $a_{3} = 3a_{2} + 2a_{1}$ $\newline$ $a_{3} = 3(2) + 2(5)$ $\newline$ $a_{3} = 6 + 10$ $\newline$ $a_{3} = 16$
Find $a_{4}$ Value: Find the value of $a_{4}$ using the recursive formula and the values of $a_{2}$ and $a_{3}$ . Now we use the formula $a_{n}=3a_{n-1}+2a_{n-2}$ with $n=4$ , $a_{3}=16$ , and $a_{2}=2$ . $a_{4} = 3a_{3} + 2a_{2}$ $a_{4} = 3(16) + 2(2)$ $a_{4}$ $0$ $a_{4}$ $1$