\begin{cases} f(1)=-3 f(n)=2 \cdot f(n-1) + 1 \end{cases}
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Mathematics, 09.12.2020 15:20 shanicar33500
\begin{cases} f(1)=-3 f(n)=2 \cdot f(n-1) + 1 \end{cases}
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f(1)=−3
f(n)=2⋅f(n−1)+1
f(2)=f(2)=f, left parenthesis, 2, right parenthesis, equals
Answers: 1
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