If f(0)=0, f(1)=1, f(2)=2 and f(x)=f(x-2)+f(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$. We calculate the values step by step: For $x=3: f

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$. We calculate the values step by step: For $x=3: f(3)=f(1)+f(0)=1+0=1$. For $x=4: f(4)=f(2)+f(1)=2+1=3$. For $x=5: f(5)=f(3)+f(2)=1+2=3$. For $x=6: f(6)=f(4)+f(3)=3+1=4$. For $x=7: f(7)=f(5)+f(4)=3+3=6$. For $x=8: f(8)=f(6)+f(5)=4+3=7$. For $x=9: f(9)=f(7)+f(6)=6+4=10$. Thus, $f(9)=10$.

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$, then $f(9)$ is equal to

Similar Questions

One mole of calcium phosphide on reaction with excess of water gives

Which of the following components of cement is present in the highest amount?

What do bio-pesticides include?

If $\int \frac{dx}{x^2+2x+2} = f(x) + c$, then $f(x)$ is equal to :

$A$ hollow metal sphere of radius $5\, cm$ is charged so that the potential on its surface is $10\, V$. The potential at the centre of the sphere is.....$V$

Vedclass Test Series

Exam Paper Generator

Online Exam Module