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$Consider a real-valued function f (x) satisfying 2 f (x y)$ $= (f (x))^{y} + (f (y)^{x} \forall x, y \in R and f (1) = a where a \neq 1 . then$ $(a - 1) \sum_{l = 1}^{n} f (i) =$

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Correct option is D

We have 2f(xy)=(f(x))y+(f(y))x Replacing y by 1, we get 2f(x)=f(x)+(f(1))x⇒f(x)=ax⇒ ∑i=1n f(i)=a+a2+⋯+an=an+1−aa−1⇒ (a−1)∑i=1n f(i)=an+1−a

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Consider a real-valued function f(x) satisfying 2f(xy) =(f(x))y+f(y)x ∀x,y∈R and f(1)=a where a≠1. then (a−1)∑l=1n f(i)=

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$Consider a real-valued function f (x) satisfying 2 f (x y)$ $= (f (x))^{y} + (f (y)^{x} \forall x, y \in R and f (1) = a where a \neq 1 . then$ $(a - 1) \sum_{l = 1}^{n} f (i) =$