If a_(1)=4,a_(2)=1 and a_(n)=a_(n-1)+3a_(n-2) then find the value of a_(6). Answer:

Given Sequence and Formula: We are given the first two terms of the sequence: a_(1) = 4 and a_(2) = 1. We also have the recursive formula a_(n) = a_(n-1) + 3a_(n-2). To find a_(6), we need to find the terms a_(3), a_(4), a_(5), and then a_(6) using the recursive formula.Find a_(3): Let's find a_(3) using the recursive formula: a_(3) = a_(2) + 3a_(1) = 1 + 3(4) = 1 + 12 = 13.Find a_(4): Now, let's find a_(4) using the recursive formula: a_(4) = a_(3) + 3a_(2) = 13 + 3(1) = 13 + 3 = 16.Find a_(5): Next, we find a_(5) using the recursive formula: a_(5) = a_(4) + 3a_(3) = 16 + 3(13) = 16 + 39 = 55.Find a_(6): Finally, we find a_(6) using the recursive formula: a_(6) = a_(5) + 3a_(4) = 55 + 3(16) = 55 + 48 = 103.

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If
a_(1)=4,a_(2)=1 and
a_(n)=a_(n-1)+3a_(n-2) then find the value of
a_(6).
Answer:

If $a_{1}=4, a_{2}=1$ and $a_{n}=a_{n-1}+3 a_{n-2}$ then find the value of $a_{6}$ . $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

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Q. If

a_{1}=4, a_{2}=1

and

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

then find the value of

a_{6}

\newline

Answer:

Given Sequence and Formula: We are given the first two terms of the sequence: $a_1 = 4$ and $a_2 = 1$ . We also have the recursive formula $a_n = a_{n-1} + 3a_{n-2}$ . To find $a_6$ , we need to find the terms $a_3$ , $a_4$ , $a_5$ , and then $a_6$ using the recursive formula.
Find $a_{3}$ : Let's find $a_{3}$ using the recursive formula: $a_{3} = a_{2} + 3a_{1} = 1 + 3(4) = 1 + 12 = 13$ .
Find $a_{4}$ : Now, let's find $a_{4}$ using the recursive formula: $a_{4} = a_{3} + 3a_{2} = 13 + 3(1) = 13 + 3 = 16$ .
Find $a_{5}$ : Next, we find $a_{5}$ using the recursive formula: $a_{5} = a_{4} + 3a_{3} = 16 + 3(13) = 16 + 39 = 55$ .
Find $a_{6}$ : Finally, we find $a_{6}$ using the recursive formula: $a_{6} = a_{5} + 3a_{4} = 55 + 3(16) = 55 + 48 = 103$ .