{:[{[h(1)=-35],[h(n)=h(n-1)*2]:}],[h(3)=]:}

Given initial condition: We are given the initial condition: h(1) = -35 And the recursive formula: h(n) = h(n - 1) * 2 We need to find the value of h(3).Finding h(2): First, let's find the value of h(2) using the recursive formula: h(2) = h(2 - 1) * 2 h(2) = h(1) * 2 h(2) = -35 * 2 h(2) = -70Finding h(3): Now, let's use the value of h(2) to find h(3): h(3) = h(3 - 1) * 2 h(3) = h(2) * 2 h(3) = -70 * 2 h(3) = -140

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{:[{[h(1)=-35],[h(n)=h(n-1)*2]:}],[h(3)=]:}

$\begin{array}{l}\left\{\begin{array}{l}h(1)=-35 \\ h(n)=h(n-1) \cdot 2\end{array}\right. \\ h(3)=\end{array}$

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Algebra 1

Evaluate recursive formulas for sequences

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\begin{array}{l}\left\{\begin{array}{l}h(1)=-35 \\ h(n)=h(n-1) \cdot 2\end{array}\right. \\ h(3)=\end{array}

Given initial condition: We are given the initial condition: $\newline$ $h(1) = -35$ $\newline$ And the recursive formula: $\newline$ $h(n) = h(n - 1) \times 2$ $\newline$ We need to find the value of $h(3)$ .
Finding $h(2)$ : First, let's find the value of $h(2)$ using the recursive formula: $\newline$ $h(2) = h(2 - 1) \times 2$ $\newline$ $h(2) = h(1) \times 2$ $\newline$ $h(2) = (-35) \times 2$ $\newline$ $h(2) = -70$
Finding h( $3$ ): Now, let's use the value of h( $2$ ) to find h( $3$ ): $\newline$ h( $3$ ) = h( $3$ - $1$ ) $\cdot$ $2$ $\newline$ h( $3$ ) = h( $2$ ) $\cdot$ $2$ $\newline$ h( $3$ ) = $-70$ $\cdot$ $2$ $\newline$ h( $3$ ) = $-140$