Paragraph: Let f(x) be a function satisfying f(x+y)=f(x) f(y) for...

Given that $f(x+y)=f(x) \cdot f(y)$ and $f(1)=2$$\begin{aligned} & \therefore \quad f(2)=f(1+1)=f(1) f(1)=2.2=4 \\ & f(3)=f(1+2)=f(1) \cdot f(2)=2.4=8 \\ & f(4)=f(1+3)=f(1) \cdot f(3)=2.8=16, \text{ So on } \end{aligned}$Now, $\sum_{x=2}^n f(x)=2044\begin{aligned} & f(2)+f(3)+f(4)+\ldots+f(n)=2044 \\ & 4+8+16 \ldots(n-1) \text{ terms }=2044 \\ & \frac{4\left(2^{n-1}-1\right)}{2-1}=2044 \Rightarrow 2^{n-1}-1=511 \\ & \Rightarrow 2^{\mathrm{n}-1}=512=2^9 \\ & \Rightarrow n=10 \end{aligned}$$$