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If
a_(1)=1 and
a_(n)=3a_(n-1)+2 then find the value of
a_(5).
Answer:

If $a_{1}=1$ and $a_{n}=3 a_{n-1}+2$ then find the value of $a_{5}$ . $\newline$ Answer:

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Find trigonometric ratios using multiple identities

Full solution

Q. If

a_{1}=1

and

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

then find the value of

a_{5}

.

\newline

Answer:

Given Information: We are given the first term of the sequence, $a_{1} = 1$ , and the recursive formula $a_{n} = 3a_{n-1} + 2$ . 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: $\newline$ $a_{2} = 3a_{1} + 2 = 3(1) + 2 = 3 + 2 = 5$ .
Find $a_{3}$ : Next, we find $a_{3}$ using the value of $a_{2}$ : $a_{3} = 3a_{2} + 2 = 3(5) + 2 = 15 + 2 = 17$ .
Find $a_{4}$ : Now, we find $a_{4}$ using the value of $a_{3}$ : $\newline$ $a_{4} = 3a_{3} + 2 = 3(17) + 2 = 51 + 2 = 53$ .
Find $a_{5}$ : Finally, we find $a_{5}$ using the value of $a_{4}$ : $\newline$ $a_{5} = 3a_{4} + 2 = 3(53) + 2 = 159 + 2 = 161$ .

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