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

If $a_{1}=0, a_{2}=3$ and $a_{n}=2 a_{n-1}-a_{n-2}$ then find the value of $a_{4}$ . $\newline$ Answer:

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Write and solve direct variation equations

Full solution

Q. If

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

and

a_{n}=2 a_{n-1}-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 $a_{n}=2a_{n-1}-a_{n-2}$ , which means each term is calculated by doubling the previous term and subtracting the term before that. We are given $a_{1}=0$ and $a_{2}=3$ as initial conditions.
Find $a_{3}$ : Find the value of $a_{3}$ using the recursive formula. $\newline$ Using the formula $a_{n}=2a_{n-1}-a_{n-2}$ , we substitute $n=3$ to find $a_{3}$ . $\newline$ $a_{3} = 2a_{3-1} - a_{3-2}$ $\newline$ $a_{3} = 2a_{2} - a_{1}$ $\newline$ $a_{3} = 2\times3 - 0$ $\newline$ $a_{3} = 6 - 0$ $\newline$ $a_{3} = 6$
Find $a_{4}$ : Find the value of $a_{4}$ using the recursive formula. $\newline$ Now that we have $a_{3}$ , we can use the formula $a_{n}=2a_{n-1}-a_{n-2}$ to find $a_{4}$ . $\newline$ $a_{4} = 2a_{4-1} - a_{4-2}$ $\newline$ $a_{4} = 2a_{3} - a_{2}$ $\newline$ $a_{4} = 2\times 6 - 3$ $\newline$ $a_{4} = 12 - 3$ $\newline$ $a_{4} = 9$

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