Let f ( x ) + f ( x + 2 ) = f ( x + 1 ) − f ( x − 1 ) for x ∈ N , and f ( 1 ) = 1986 . Then the sum of the digits of f ( 1 ) + f ( 2 ) + ⋯ + f ( 1985 ) is:

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Let $f (x) + f (x + 2) = f (x + 1) - f (x - 1)$ for $x \in N$ , and $f (1) = 1986$ . Then the sum of the digits of $f (1) + f (2) + \dots + f (1985)$ is:

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Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

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