Let $f:N\to R$ be a function such that f(x+y)=2 f(x)f(y) for natural numbers x and y. If f(1)=2, then the value of $\alpha$ for which
$\sum _{k=1}^{10}f(\alpha +k)=\frac{512}{3}\left(2^{20}-1\right)$
holds, is

$$\begin{gathered} f:N\to R,f(x+y)=2f\left(x\right)f\left(y\right) ...\left(1\right) \\ f\left(1\right)=2\text{, } \\ \sum _{\mathrm{k}=1}^{10}\mathrm{f}(\alpha +\mathrm{k})=2\mathrm{f}\left(\alpha \right)\sum _{\mathrm{k}=1}^{10}\mathrm{f}\left(\mathrm{k}\right) \\ =2f\left(\alpha \right)\left(f\right(1)+f(2)+\dots ..+f(10\left)\right) \\ \text{ From (1) } \\ f\left(2\right)=2f^2\left(1\right)=2^3 \\ f\left(3\right)=2f\left(2\right)f\left(1\right)=2^5 \\ . \\ . \\ . \\ f\left(10\right)=2^9{f}^{10}\left(1\right)=2^{19} \\ f\left(\alpha \right)=2^{2\alpha -1};\alpha \in N \\ \text{ from }\left(2\right) \\ \sum _{\mathrm{k}=1}^{10}\mathrm{f}(\alpha +\mathrm{k})=2\left(2^{2\alpha -1}\right)\left(2+2^3+2^5+\dots .+2^{19}\right) \\ \frac{512}{3}\left(2^{20}-1\right)=2^{2\alpha }\left(2\frac{\left(2^{20}-1\right)}{3}\right) \\ \text{ Hence }\alpha =4 \end{gathered}$$