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 $$\sum_{k=1}^{10} f(\alpha + k) = \frac{512}{3}2^{20} - 1$$ holds, is

We need to find $$\alpha$$ such that $$\sum_{k=1}^{10}f(\alpha + k)=\frac{512}{3}(2^{20}-1)\,.$$

First, observe that $$f(x+y)=2f(x)f(y)$$ with $$f(1)=2$$. Substituting $$x=y=1$$ gives $$f(2)=2f(1)^2=2\times4=8=2^3$$, and then for $$x=1,y=2$$ we get $$f(3)=2f(1)f(2)=2\times2\times8=32=2^5$$. Likewise, $$x=1,y=3$$ yields $$f(4)=2f(1)f(3)=2\times2\times32=128=2^7$$, so by induction the pattern is $$f(n)=2^{2n-1}$$, which indeed satisfies $$f(x+y)=2^{2(x+y)-1}=2\cdot2^{2x-1}\cdot2^{2y-1}=2f(x)f(y)\,. $$

Since $$f(n)=2^{2n-1}$$, the sum becomes $$\sum_{k=1}^{10}f(\alpha + k)=\sum_{k=1}^{10}2^{2(\alpha+k)-1}=2^{2\alpha-1}\sum_{k=1}^{10}2^{2k}=2^{2\alpha-1}\sum_{k=1}^{10}4^k$$ which simplifies to $$2^{2\alpha-1}\cdot\frac{4(4^{10}-1)}{4-1}=2^{2\alpha-1}\cdot\frac{4(2^{20}-1)}{3}\,.$$

Equating this to $$\frac{512}{3}(2^{20}-1)$$ gives $$2^{2\alpha-1}\cdot\frac{4(2^{20}-1)}{3}=\frac{512}{3}(2^{20}-1)\,, $$ hence $$2^{2\alpha-1}\cdot4=512\,, $$ or $$2^{2\alpha-1}\cdot2^2=2^9\,, $$ so $$2^{2\alpha+1}=2^9$$ and therefore $$2\alpha+1=9\implies\alpha=4\,. $$

Thus the correct answer is $$4$$.