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

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

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Find trigonometric ratios using multiple identities

Full solution

Q. If

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

and

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

then find the value of

a_{4}

.

\newline

Answer:

Given Sequence and Formula: We are given the first two terms of the sequence: $a_{1}=0$ and $a_{2}=2$ . We also have the recursive formula $a_{n}=a_{n-1}+3a_{n-2}$ . To find $a_{4}$ , we first need to find $a_{3}$ using the recursive formula.
Calculate $a_{3}$ : Using the recursive formula, let's calculate $a_{3}$ : $a_{3} = a_{3-1} + 3a_{3-2}$ $= a_{2} + 3a_{1}$ $= 2 + 3(0)$ $= 2 + 0$ $= 2$ We have found that $a_{3} = 2$ .
Calculate $a_{4}$ : Now we can use the values of $a_{2}$ and $a_{3}$ to find $a_{4}$ using the same recursive formula: $\newline$ $a_{4} = a_{4-1} + 3a_{4-2}$ $\newline$ $= a_{3} + 3a_{2}$ $\newline$ $= 2 + 3(2)$ $\newline$ $= 2 + 6$ $\newline$ $= 8$ $\newline$ We have found that $a_{4} = 8$ .

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