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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 ______

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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