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

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

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Full solution

Q. {h(1)=−2h(2)=1h(n)=h(n−2)+h(n−1)h(3)= \begin{array}{l}\left\{\begin{array}{l}h(1)=-2 \\ h(2)=1 \\ h(n)=h(n-2)+h(n-1)\end{array}\right. \\ h(3)=\end{array}
  1. Given initial conditions: We are given the initial conditions of the recursive sequence:\newlineh(1)=−2h(1) = -2\newlineh(2)=1h(2) = 1\newlineAnd the recursive formula:\newlineh(n)=h(n−2)+h(n−1)h(n) = h(n-2) + h(n-1)\newlineWe need to find the value of h(3)h(3).
  2. Recursive formula: Using the recursive formula, we can find h(3)h(3) by adding h(1)h(1) and h(2)h(2):
    h(3)=h(1)+h(2)h(3) = h(1) + h(2)
    h(3)=(−2)+1h(3) = (-2) + 1
    h(3)=−1h(3) = -1

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