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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} ⎩








f(1)=−3
f(n)=2⋅f(n−1)+1


f(2)=f(2)=f, left parenthesis, 2, right parenthesis, equals

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\begin{cases} f(1)=-3 f(n)=2 \cdot f(n-1) + 1 \end{cases} ⎩




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