{[g(1)=14],[g(n)=g(n-1)-4]:} Find an explicit formula for g(n). g(n)=

Initialize g(1): g(1) = 14, and g(n) = g(n-1) - 4. To find g(2), we substitute n=2 into the recursive formula. g(2) = g(2-1) - 4 = g(1) - 4 = 14 - 4 = 10.Find g(2): Now, let's find g(3) to see the pattern. g(3) = g(3-1) - 4 = g(2) - 4 = 10 - 4 = 6.Find g(3): We can see that each step we subtract 4 from the previous g(n). So, for g(n), we subtract 4 (n-1) times from g(1). g(n) = g(1) - 4(n-1).General formula for g(n): Substitute g(1) = 14 into the formula. g(n) = 14 - 4(n-1).Substitute g(1): Simplify the formula. g(n) = 14 - 4n + 4.Simplify the formula: Combine like terms. g(n) = 18 - 4n.

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{[g(1)=14],[g(n)=g(n-1)-4]:}
Find an explicit formula for
g(n).

g(n)=

$\left\{\begin{array}{l} g(1)=14 \\ g(n)=g(n-1)-4 \end{array}\right.$ $\newline$ Find an explicit formula for $g(n)$ . $\newline$ $g(n)=$

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Algebra 2

Write a formula for an arithmetic sequence

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\left\{\begin{array}{l} g(1)=14 \\ g(n)=g(n-1)-4 \end{array}\right.

\newline

Find an explicit formula for

g(n)

\newline

g(n)=

Initialize $g(1)$ : $g(1) = 14$ , and $g(n) = g(n-1) - 4$ . To find $g(2)$ , we substitute $n=2$ into the recursive formula. $\newline$ $g(2) = g(2-1) - 4 = g(1) - 4 = 14 - 4 = 10$ .
Find $g(2)$ : Now, let's find $g(3)$ to see the pattern. $\newline$ $g(3) = g(3-1) - 4 = g(2) - 4 = 10 - 4 = 6$ .
Find $g(3)$ : We can see that each step we subtract $4$ from the previous $g(n)$ . So, for $g(n)$ , we subtract $4$ $(n-1)$ times from $g(1)$ . $\newline$ $g(n) = g(1) - 4(n-1)$ .
General formula for $g(n)$ : Substitute $g(1) = 14$ into the formula. $\newline$ $g(n) = 14 - 4(n-1)$ .
Substitute $g(1)$ : Simplify the formula. $g(n) = 14 - 4n + 4.$
Simplify the formula: Combine like terms. $g(n) = 18 - 4n$ .