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The function f(n) satisfies f(1) = 1 and f(2n + 1) = f(n) + 2 for n >= 0.  Find f(15).

 Jun 3, 2021

Best Answer 

 #1
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The function \(f(n)\) satisfies \(f(1)=1\) and \(f(2n+1)=f(n)+2\) for \(n \ge 0\).
Find \(f(15)\).

 

\(\begin{array}{|lrcll|} \hline n=1:& f(2*1+1) &=& f(1) + 2 \quad | \quad f(1)=1 \\ &f(3) &=& 1+2 \\ &f(3)&=& 3 \\ \hline n=3:& f(2*3+1) &=& f(3) + 2 \quad | \quad f(3)=3 \\ &f(7) &=& 3+2 \\ &f(7)&=& 5 \\ \hline n=7:& f(2*7+1) &=& f(7) + 2 \quad | \quad f(7)=5 \\ &f(15) &=& 5+2 \\ &\mathbf{ f(15) } &=& \mathbf{7} \\ \hline \end{array}\)

 

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 Jun 3, 2021
 #1
avatar+26213 
+2
Best Answer

The function \(f(n)\) satisfies \(f(1)=1\) and \(f(2n+1)=f(n)+2\) for \(n \ge 0\).
Find \(f(15)\).

 

\(\begin{array}{|lrcll|} \hline n=1:& f(2*1+1) &=& f(1) + 2 \quad | \quad f(1)=1 \\ &f(3) &=& 1+2 \\ &f(3)&=& 3 \\ \hline n=3:& f(2*3+1) &=& f(3) + 2 \quad | \quad f(3)=3 \\ &f(7) &=& 3+2 \\ &f(7)&=& 5 \\ \hline n=7:& f(2*7+1) &=& f(7) + 2 \quad | \quad f(7)=5 \\ &f(15) &=& 5+2 \\ &\mathbf{ f(15) } &=& \mathbf{7} \\ \hline \end{array}\)

 

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heureka Jun 3, 2021

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