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

Given information: We are given the first term of the sequence, a_(1)=4, and the recursive formula for the sequence, a_(n+1)=2a_(n)+4. To find a_(3), we first need to find a_(2). Using the recursive formula, we substitute n=1 to find a_(2): a_(2) = 2a_(1) + 4 a_(2) = 2*4 + 4 a_(2) = 8 + 4 a_(2) = 12Find a_(2): Now that we have a_(2)=12, we use the recursive formula again to find a_(3). We substitute n=2 into the formula: a_(3) = 2a_(2) + 4 a_(3) = 2*12 + 4 a_(3) = 24 + 4 a_(3) = 28

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

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

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Precalculus

Find trigonometric ratios using multiple identities

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Q. If

a_{1}=4

and

a_{n+1}=2 a_{n}+4

then find the value of

a_{3}

\newline

Answer:

Given information: We are given the first term of the sequence, $a_{1}=4$ , and the recursive formula for the sequence, $a_{n+1}=2a_{n}+4$ . To find $a_{3}$ , we first need to find $a_{2}$ . $\newline$ Using the recursive formula, we substitute $n=1$ to find $a_{2}$ : $\newline$ $a_{2} = 2a_{1} + 4$ $\newline$ $a_{2} = 2\times4 + 4$ $\newline$ $a_{2} = 8 + 4$ $\newline$ $a_{2} = 12$
Find $a_{2}$ : Now that we have $a_{2}=12$ , we use the recursive formula again to find $a_{3}$ . We substitute $n=2$ into the formula: $\newline$ $a_{3} = 2a_{2} + 4$ $\newline$ $a_{3} = 2\times 12 + 4$ $\newline$ $a_{3} = 24 + 4$ $\newline$ $a_{3} = 28$