The function f(n) satisfies f(1) = 1 and f(2n + 1) = f(n) + 2 for n >= 0. Find f(15).
The function f(n) satisfies f(1)=1 and f(2n+1)=f(n)+2 for n≥0.Find f(15).
n=1:f(2∗1+1)=f(1)+2|f(1)=1f(3)=1+2f(3)=3n=3:f(2∗3+1)=f(3)+2|f(3)=3f(7)=3+2f(7)=5n=7:f(2∗7+1)=f(7)+2|f(7)=5f(15)=5+2f(15)=7