Question

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

• A. $nf\left(1\right)$
• B. $\left\{f\right(1{)}^n\}$
• C. $0$
• D. None of these

Answer 1

The correct option is A $nf\left(1\right)$

$\Rightarrow$$f(x-1)+f(x+1)=2f\left(x\right)$ ...(it is given that $f\left(0\right)=0$)

Substituting $x=1$, we get

$\Rightarrow$$f\left(0\right)+f\left(2\right)=2f\left(1\right)$

$\Rightarrow$$f\left(2\right)=2f\left(1\right)$ ...$\left(f\right(0)=0)$

Substituting $x=2$

$\Rightarrow$$f\left(1\right)+f\left(3\right)=2f\left(2\right)$

$\Rightarrow$$f\left(1\right)+f\left(3\right)=4f\left(1\right)$ ...($f\left(2\right)=2f\left(1\right)$)

$\Rightarrow$$f\left(3\right)=3f\left(1\right)$

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Therefore $f\left(n\right)=n\left(f\right(1\left)\right)$