f ( x + 1 ) = ( − 1 ) x + 1 x − 1 f ( x ) for x ∈ N and f ( 1 ) = f ( 1986 ) . Then sum of digits of ( f ( 1 ) + f ( 2 ) + …… .. + f ( 1985 )) is

∑ x = 1 1985 f ( x + 1 ) = ∑ x = 1 1985 ( − 1 ) x + 1 x − 2 ∑ x = 1 1985 f ( x )

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$f (x + 1) = (- 1)^{x + 1} x - 1 f (x)$ for $x \in N$ and $f (1) = f (1986)$ . Then sum of digits of $(f (1) + f (2) + \dots\dots .. + f (1985))$ is

Answer:

Solution:

$\sum_{x = 1}^{1985} f (x + 1) = \sum_{x = 1}^{1985} (- 1)^{x + 1} x - 2 \sum_{x = 1}^{1985} f (x)$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

⏱ 2mℹ️ Source: QB

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