If $$f\left( {x + 1} \right) + f\left( {x - 1} \right) = 2f\left( x \right)$$      and $$f\left( 0 \right) = 0$$   then $$f\left( n \right),\,n\, \in \,N,$$    is :

Solution :
$$ \Rightarrow f\left( {x - 1} \right) + f\left( {x + 1} \right) = 2f\left( x \right)....$$        (it is given that $${f\left( 0 \right) = 0}$$  )
$$\eqalign{ & {\text{Substituting }}x = 1,{\text{ we get}} \cr & \Rightarrow f\left( 0 \right) + f\left( 2 \right) = 2f\left( 1 \right) \cr & \Rightarrow f\left( 2 \right) = 2f\left( 1 \right)....\left( {f\left( 0 \right) = 0} \right) \cr & {\text{Substituting }}x = 2,{\text{ we get}} \cr & \Rightarrow f\left( 1 \right) + f\left( 3 \right) = 2f\left( 2 \right) \cr & \Rightarrow f\left( 1 \right) + f\left( 3 \right) = 4f\left( 1 \right)....\left( {f\left( 2 \right) = 2f\left( 1 \right)} \right) \cr & \Rightarrow f\left( 3 \right) = 3f\left( 1 \right) \cr & {\text{:}} \cr & {\text{:}} \cr & {\text{Therefore }}f\left( n \right) = n\left( {f\left( 1 \right)} \right) \cr} $$