If a_(1)=0,a_(2)=1 and a_(n)=3a_(n-1)+3a_(n-2) then find the value of a_(6). Answer:

Initialize Sequence Values: To find the value of a_(6), we need to use the given recursive formula a_(n)=3a_(n-1)+3a_(n-2) and the initial conditions a_(1)=0 and a_(2)=1 to calculate the sequence values up to n=6.Calculate a_(3): First, we calculate a_(3) using the recursive formula: a_(3) = 3a_(2) + 3a_(1) = 3*1 + 3*0 = 3 + 0 = 3.Calculate a_(4): Next, we calculate a_(4): a_(4) = 3a_(3) + 3a_(2) = 3*3 + 3*1 = 9 + 3 = 12.Calculate a_(5): Then, we calculate a_(5): a_(5) = 3a_(4) + 3a_(3) = 3*12 + 3*3 = 36 + 9 = 45.Calculate a_(6): Finally, we calculate a_(6): a_(6) = 3a_(5) + 3a_(4) = 3*45 + 3*12 = 135 + 36 = 171.

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If
a_(1)=0,a_(2)=1 and
a_(n)=3a_(n-1)+3a_(n-2) then find the value of
a_(6).
Answer:

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

and

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

then find the value of

a_{6}

\newline

Answer:

Initialize Sequence Values: To find the value of $a_{6}$ , we need to use the given recursive formula $a_{n}=3a_{n-1}+3a_{n-2}$ and the initial conditions $a_{1}=0$ and $a_{2}=1$ to calculate the sequence values up to $n=6$ .
Calculate $a_{3}$ : First, we calculate $a_{3}$ using the recursive formula: $\newline$ $a_{3} = 3a_{2} + 3a_{1} = 3\times 1 + 3\times 0 = 3 + 0 = 3$ .
Calculate $a_{4}$ : Next, we calculate $a_{4}$ : $a_{4} = 3a_{3} + 3a_{2} = 3\times3 + 3\times1 = 9 + 3 = 12$ .
Calculate $a_{5}$ : Then, we calculate $a_{5}$ : $a_{5} = 3a_{4} + 3a_{3} = 3\times12 + 3\times3 = 36 + 9 = 45$ .
Calculate $a_{6}$ : Finally, we calculate $a_{6}$ : $a_{6} = 3a_{5} + 3a_{4} = 3\times45 + 3\times12 = 135 + 36 = 171$ .