Let f : N $\to$ R be a function such that $f(x + y) = 2f(x)f(y)$ for natural numbers x and y. If f(1) = 2, then the value of $\alpha$ for which

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

holds, is :

<p>Given,</p> <p>$f(x + y) = 2f(x)f(y)$</p> <p>and $f(1) = 2$</p> <p>For x = 1 and y = 1,</p> <p>$f(1 + 1) = 2f(1)f(1)$</p> <p>$\Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$</p> <p>For x = 1, y = 2,</p> <p>$f(1 + 2) = 2f(1)y(2)$</p> <p>$\Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$</p> <p>For x = 1, y = 3,</p> <p>$f(1 + 3) = 2f(1)f(3)$</p> <p>$\Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$</p> <p>For x = 1, y = 4,</p> <p>$f(1 + 4) = 2f(1)f(4)$</p> <p>$\Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$ ..... (1)</p> <p>Also given</p> <p>$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)}$</p> <p>$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)$$</p> <p>$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$$</p> <p>This represent a G.P with first term = 2⁹ and common ratio = 2²</p> <p>$\therefore$ First term $= f(\alpha + 1) = {2^9}$ ..... (2)</p> <p>From equation (1), $f(5) = {2^9}$</p> <p>$\therefore$ From (1) and (2), we get</p> <p>$f(\alpha + 1) = {2^9} = f(5)$</p> <p>$\Rightarrow f(\alpha + 1) = f(5)$</p> <p>$\Rightarrow f(\alpha + 1) = f(4 + 1)$</p> <p>Comparing both sides we get,</p> <p>$\alpha = 4$</p>