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

{h(1)=−35h(n)=h(n−1)⋅2h(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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Q. {h(1)=−35h(n)=h(n−1)⋅2h(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}
  1. Given initial condition: We are given the initial condition:\newlineh(1)=−35h(1) = -35\newlineAnd the recursive formula:\newlineh(n)=h(n−1)×2h(n) = h(n - 1) \times 2\newlineWe need to find the value of h(3)h(3).
  2. Finding h(2)h(2): First, let's find the value of h(2)h(2) using the recursive formula:\newlineh(2)=h(2−1)×2h(2) = h(2 - 1) \times 2\newlineh(2)=h(1)×2h(2) = h(1) \times 2\newlineh(2)=(−35)×2h(2) = (-35) \times 2\newlineh(2)=−70h(2) = -70
  3. Finding h(33): Now, let's use the value of h(22) to find h(33):\newlineh(33) = h(33 - 11) ⋅\cdot 22\newlineh(33) = h(22) ⋅\cdot 22\newlineh(33) = −70-70 ⋅\cdot 22\newlineh(33) = −140-140

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