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If $f(x)$ is a real valued function satisfying $f(x+1)+f(x-1)=2f(x)$ $\forall x\in R$ and $f(0)=0$, then for any $n\in N,\:f(n)$ = ?

$\begin{array}{1 1} nf(1) \\ [f(1)]^n \\ 0 \\ none\;of\;these \end{array}$

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Given $f(x+1)+f(x-1)=2f(x)\:\forall x\in R$
Put $x=1$ $\Rightarrow\:f(2)+f(0)=2f(1)$
put $x=2\:\Rightarrow\:f(3)+f(1)=2f(2)=4f(1)$
Similarly by putting $x=3$ we get f(4)=4f(1)..........
answered May 7, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1

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