If f(x+1)+f(x−1)=2f(x) and f(0)=0, then f(n),(n∈N) is

put x=1 then f(2)+f(0)=2f(1)⇒f(2)=2f(1)put x=2,f(3)=3f(1)...

If f(x+1)+f(x−1)=2f(x) and f(0)=0, then f(n),(n∈N) is

put x=1 then f(2)+f(0)=2f(1)⇒f(2)=2f(1)put x=2,f(3)=3f(1)...

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

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$n f (1)$

${[f (1)]}^{n}$

$f (1)$

answer is A.

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$put x = 1 then f (2) + f (0) = 2 f (1) \Rightarrow f (2) = 2 f (1)$

$put x = 2, f (3) = 3 f (1) ...$

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If f(x+1)+f(x−1)=2f(x) and f(0)=0, then f(n),(n∈N) is

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