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

If $a_{1}=0, a_{2}=3$ and $a_{n}=3 a_{n-1}-a_{n-2}$ 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}=3 a_{n-1}-a_{n-2}

then find the value of

a_{4}

.

\newline

Answer:

Find $a_{3}$ : Use the given recursive formula to find $a_{3}$ . The recursive formula is $a_{n}=3a_{n-1}-a_{n-2}$ . We know $a_{1}=0$ and $a_{2}=3$ . Substitute $n=3$ into the formula to find $a_{3}$ . $a_{3} = 3a_{3-1} - a_{3-2}$ $a_{3} = 3a_{2} - a_{1}$ $a_{3} = 3\times3 - 0$ $a_{3}$ $0$ $a_{3}$ $1$
Find $a_{4}$ : Use the recursive formula to find $a_{4}$ . Now that we have $a_{3}$ , we can use the formula again to find $a_{4}$ . $a_{4} = 3a_{4-1} - a_{4-2}$ $a_{4} = 3a_{3} - a_{2}$ $a_{4} = 3\times9 - 3$ $a_{4} = 27 - 3$ $a_{4} = 24$

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