If `f(x+1)+f(x-1)=2f(x)a n df(0),=0,` then `f(n),n in N ,` is `nf(1)` (b) `{f(1)}^n` `0` (d) none of these

Options

A. `n f(1)`
B. `{f(1)}^(n)`
C. 0
D. none of these

Answer

A

Explanation

Putting `x=1, f(2)+f(0)=2f(1) or f(2)=2f(1).`
Putting `x=2, f(3)+f(1)=2f(2)`
` or f(3)=2xx2f(1)-f(1)=3f(1)` and so on.
` :. f(n)=n f(1), " for " n=1,2, …, n`
`f(n+1)+f(n-1)=2f(n)`
` or f(n+1)+(n-1)f(1)=2nf(1)`
`or f(n+1)=(n+1)f(1)`