Given,

$$f(x + y) = 2f(x)f(y)$$

and $$f(1) = 2$$

For x = 1 and y = 1,

$$f(1 + 1) = 2f(1)f(1)$$

$$ \Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$$

For x = 1, y = 2,

$$f(1 + 2) = 2f(1)y(2)$$

$$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$$

For x = 1, y = 3,

$$f(1 + 3) = 2f(1)f(3)$$

$$ \Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$$

For x = 1, y = 4,

$$f(1 + 4) = 2f(1)f(4)$$

$$ \Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$$ ..... (1)

Also given

$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $$

$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)$$

$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$$

This represent a G.P with first term = 2⁹ and common ratio = 2²

$$\therefore$$ First term $$ = f(\alpha + 1) = {2^9}$$ ..... (2)

From equation (1), $$f(5) = {2^9}$$

$$\therefore$$ From (1) and (2), we get

$$f(\alpha + 1) = {2^9} = f(5)$$

$$ \Rightarrow f(\alpha + 1) = f(5)$$

$$ \Rightarrow f(\alpha + 1) = f(4 + 1)$$

Comparing both sides we get,

$$\alpha = 4$$