If $\quad f(0)=0, f(1)=1, f(2)=2 \quad$ and $f(x)=f(x-2)+f(x-3)$ for $x=3,4,5, \ldots$, then $f(9)$ is equal…
If $\quad f(0)=0, f(1)=1, f(2)=2 \quad$ and $f(x)=f(x-2)+f(x-3)$ for $x=3,4,5, \ldots$, then $f(9)$ is equal to
12
13
14
10
Solution
$f(0)=0, f(1)=1, f(2)=2$
Given,
$f(x)=f(x-2)+f(x-3), x=3,4,5, \ldots$
The given function is known as "Reccurrence function".
put $x=3, f(3)=f(1)+f(0)$
$=1+0=1$
put $x=4, f(4)=f(2)+f(1)$
$=2+1 \Rightarrow 3$
put $x=5, f(5)=f(3)+f(2)$
$=1+2 \Rightarrow 3$
put $x=6, f(6)=f(4)+f(3)$
$=3+1 \Rightarrow 3+1 \Rightarrow 4$
put $x=7, f(7)=f(5)+f(4)$
$=3+3 \Rightarrow 6$
put $x=8, f(8)=f(6)+f(5)$
$=3+4 \Rightarrow 7$
put $x=9, f(9)=f(7)+f(6)$
$=6+4 \Rightarrow 10$
Hence, $\quad f(9)=10$