If a_(1)=1 and a_(n)=3a_(n-1)-n then find the value of a_(5). Answer:

Given Sequence and Formula: We are given the first term of the sequence, a_(1) = 1, and the recursive formula a_(n) = 3a_(n-1) - n. To find a_(5), we need to find the values of a_(2), a_(3), a_(4), and then a_(5) using the recursive formula.Find a_(2): First, let's find a_(2) using the recursive formula: a_(2) = 3a_(1) - 2 a_(2) = 3(1) - 2 a_(2) = 3 - 2 a_(2) = 1Find a_(3): Next, we find a_(3) using the value of a_(2): a_(3) = 3a_(2) - 3 a_(3) = 3(1) - 3 a_(3) = 3 - 3 a_(3) = 0Calculate a_(4): Now, we calculate a_(4) using the value of a_(3): a_(4) = 3a_(3) - 4 a_(4) = 3(0) - 4 a_(4) = 0 - 4 a_(4) = -4Find a_(5): Finally, we find a_(5) using the value of a_(4): a_(5) = 3a_(4) - 5 a_(5) = 3(-4) - 5 a_(5) = -12 - 5 a_(5) = -17

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If
a_(1)=1 and
a_(n)=3a_(n-1)-n then find the value of
a_(5).
Answer:

If $a_{1}=1$ and $a_{n}=3 a_{n-1}-n$ then find the value of $a_{5}$ . $\newline$ Answer:

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Precalculus

Find trigonometric ratios using multiple identities

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Q. If

a_{1}=1

and

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

then find the value of

a_{5}

\newline

Answer:

Given Sequence and Formula: We are given the first term of the sequence, $a_{1} = 1$ , and the recursive formula $a_{n} = 3a_{n-1} - n$ . To find $a_{5}$ , we need to find the values of $a_{2}$ , $a_{3}$ , $a_{4}$ , and then $a_{5}$ using the recursive formula.
Find $a_{2}$ : First, let's find $a_{2}$ using the recursive formula: $\newline$ $a_{2} = 3a_{1} - 2$ $\newline$ $a_{2} = 3(1) - 2$ $\newline$ $a_{2} = 3 - 2$ $\newline$ $a_{2} = 1$
Find $a_{3}$ : Next, we find $a_{3}$ using the value of $a_{2}$ : $a_{3} = 3a_{2} - 3$ $a_{3} = 3(1) - 3$ $a_{3} = 3 - 3$ $a_{3} = 0$
Calculate $a_{4}$ : Now, we calculate $a_{4}$ using the value of $a_{3}$ : $a_{4} = 3a_{3} - 4$ $a_{4} = 3(0) - 4$ $a_{4} = 0 - 4$ $a_{4} = -4$
Find $a_{5}$ : Finally, we find $a_{5}$ using the value of $a_{4}$ :
$a_{5} = 3a_{4} - 5$
$a_{5} = 3(-4) - 5$
$a_{5} = -12 - 5$
$a_{5} = -17$