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Consider a function $f$ satisfying $f(x+y)=f(x)f(y)$ where $x,y$ are positive integers, and $f(1)=2$. If $f(a+1)+f(a+2)+\ldots +f(a+n)=16(2^{n}-1)$ then $a$ is equal to ______
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Given that, $ f(x+y) = f(x) + f(y) , f(1) = 2 $

And $, f(a+1) + f(a+2) + f(a+3) + \dots + f(a+n) = 16(2^{n}-1) \quad \longrightarrow (1) $

Now, put $n=1$, we get

$ f(a+1) = 16(2-1) = 16 $

$ \Rightarrow f(a) f(1) = 16 $

$ \Rightarrow f(a) \cdot 2 = 16 $

$ \Rightarrow f(a) = 8 \quad \longrightarrow (2) $

And, put $n=2$, we get

$ f(a+1) + f(a+2) = 16(2^{2} – 1) $

$ \Rightarrow f(a) f(1) + f(a) f(2) = 48 $

$ \Rightarrow 8 \cdot 2 + 8 \cdot f(2) = 48 $

$ \Rightarrow 16 + 8 f(2) = 48 $

$ \Rightarrow 8 f(2) = 32 $

$ \Rightarrow f(2) = 4 \quad \longrightarrow (3) $

And, put $n=3$, we get

$ f(a+1) + f(a+2) + f(a+3) = 16(2^{3}-1) $

$ \Rightarrow f(a) f(1) + f(a) f(2) + f(a) f(3) = 16(7) $

$ \Rightarrow 8 \cdot2 + 8 \cdot 4 + 8 f(3) = 112 $

$ \Rightarrow 16 + 32 + 8 f(3) = 112 $

$ \Rightarrow  48 + 8 f(3) = 112 $

$ \Rightarrow 8 f(3) = 64 $

$ \Rightarrow f(3) = 8 \quad \longrightarrow (4) $

On comparing equation $(2)$ and $(4)$, we get

$ \boxed{a=3} $

Correct Answer $: 3 $
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