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F(n)=41-5ncomplete the recruisve formula of f(n)f(1)=f(n)=f(n-1)+

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\begin{gathered} f\mleft(1\mright)=36 \\ f(n)=f(n-1)-5 \end{gathered}

Step-by-step explanation

Step 1

Let


f(n)=41-5n

a) f(1)

replace the valur for n


\begin{gathered} f(n)=41-50 \\ f(1)=41-5(1)=41-5=36 \\ f(1)=36 \end{gathered}

Step 2

find f(2) and f(3)


\begin{gathered} f(2)=41-5(2)=41-10=31 \\ f(2)=31 \\ f(3)=41-5(3)=41-15=26 \end{gathered}

then, we have


\begin{gathered} f(1)=36 \\ f(2)=31 \\ f(3)=26 \end{gathered}

As you can see, the next term is the previous one minus 5,in math terms


f(n)=f(n-1)-5

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