Let f: N → R be a function such that f (x + y) = 2f (x)f(y) for...

Given $f\left(x + y\right) = 2f\left(x\right) \cdot f\left(y\right)$ & $f\left(1\right) = 2$

Now putting $x = 1$ & $y = 1$ in $f\left(x + y\right) = 2f\left(x\right) \cdot f\left(y\right)$ we get

$f\left(1 + 1\right) = 2f\left(1\right) \cdot f\left(1\right) = 2 \times {2}^{2} = {2}^{3}$

So $f\left(2\right) = {2}^{3}$, Similarly

$f\left(3\right) = {2}^{5}, f\left(4\right) = {2}^{7} . . . . .$

Now

$\mathop{∑}_{k = 1}^{10}f\left(\alpha + k\right) = \mathop{∑}_{k = 1}^{10}2f\left(\alpha\right) \cdot f\left(k\right) = \dfrac{512}{3}\left({2}^{20} - 1\right)$

$⇒ 2f\left(\alpha\right)\mathop{∑}_{k = 1}^{10}f\left(k\right) = \dfrac{512}{3}\left({2}^{20} - 1\right)$

$⇒ 2f\left(\alpha\right)\left\{f\left(1\right) + f\left(2\right) ⋯ f\left(10\right)\right\} = \dfrac{512}{3}\left({2}^{20} - 1\right)$

$⇒ 2f\left(\alpha\right)\left(2 + {2}^{3} + {2}^{5} \ldots\right) = \dfrac{512}{3}\left({2}^{20} - 1\right)$

$⇒ 2f\left(\alpha\right)\dfrac{2\left({\left({2}^{2}\right)}^{10} - 1\right)}{{2}^{2} - 1} = \dfrac{512}{3}\left({2}^{20} - 1\right)$

$⇒ 2 \times 2f\left(\alpha\right) \times \dfrac{{2}^{20} - 1}{3} = \dfrac{512}{3}\left({2}^{20} - 1\right)$

Now comparing both side we get

$4f\left(\alpha\right) = 512 ⇒ f\left(\alpha\right) = 128$

$⇒ f\left(\alpha\right) = {2}^{7} ⇒ \alpha = 4$.