Let ∑ _ k = 1 ^ 10 f(a + k) = 16( 2 ^ 10 - 1), where the function f...

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

Put $x = y = 1$

$⇒ f\left(2\right) = f\left(1\right) \cdot f\left(1\right) = {2}^{2}$

Now, put $x = 1, y = 2$

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

Now, put $x = 1, y = 3$

$⇒ f\left(4\right) = f\left(1\right) \cdot f\left(3\right) = {2}^{4}$

Similarly, $f\left(10\right) = {2}^{10}$

Now $\mathop{∑}_{k = 1}^{10}f\left(a + k\right) = 16\left({2}^{10} - 1\right)$

$⇒ \mathop{∑}_{k = 1}^{10}f\left(a\right)f\left(k\right) = 16\left({2}^{10} - 1\right)$

$⇒ f\left(a\right)\mathop{∑}_{k = 1}^{10}f\left(k\right) = 16\left({2}^{10} - 1\right)$

$⇒ f\left(a\right)\left(f\left(1\right) + f\left(2\right) \ldots + f\left(10\right)\right) = 16\left({2}^{10} - 1\right)$

$⇒ f\left(a\right)\left({2}^{1} + {2}^{2} + {2}^{3} + \ldots + {2}^{10}\right) = 16\left({2}^{10} - 1\right)$

Using the sum of $n$ terms of a geometric progression, i.e. $a + ar + a{r}^{2} + a{r}^{3} + . . . + a{r}^{n - 1} = \dfrac{a\left({r}^{n} - 1\right)}{r - 1},$ we get

$f\left(a\right)\dfrac{2\left({2}^{10} - 1\right)}{2 - 1} = 16\left({2}^{10} - 1\right)$

$⇒ f\left(a\right) = 8 = {2}^{3}$

Also, we know that $f\left(3\right) = 8$

$⇒ f\left(a\right) = f\left(3\right)$

$⇒ a = 3 .$