Given ƒ(1) = 2

and ƒ(x + y) = ƒ(x)ƒ(y)

When x = 1 and y = 1 then,

ƒ(1 + 1) = ƒ(1)ƒ(1)

$$ \Rightarrow $$ f(2) = (f(1))² = 2²

Also when x = 2 and y = 1 then,

ƒ(2 + 1) = ƒ(2)ƒ(1)

$$ \Rightarrow $$ f(3) = 2³

$$ \therefore $$ Similarly f(4) = 2⁴
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f(x) = 2^x

$$ \therefore $$ f(a + k) = 2^{a + k}

Now given,

$$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$

$$ \Rightarrow $$ $$\sum\limits_{k = 1}^{10} {{2^{a + k}}} $$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^{a + 1}} + {2^{a + 2}} + .... + {2^{a + 10}}$$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^a}\left[ {2 + {2^2} + .... + {2^{10}}} \right]$$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^a} \times {{2\left( {{2^{10}} + 1} \right)} \over {2 - 1}}$$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^a} = 8$$

$$ \Rightarrow $$ $$a$$ = 3