Question:
Let $\sum_{k=1}^{10} f(a+k)=16\left(2^{10}-1\right)$, where the function $f$ satisfies
$f(x+y)=f(x) f(y)$ for all natural numbers $x, y$ and $f(1)=2$.
Then the natural number ' $\mathrm{a}$ ' is:
Correct Option: , 4
Solution:
$\because f(x+y)=f(x) \cdot f(y)$
$\Rightarrow$ Let $f(x)=t^{x}$
$\because \mathrm{f}(1)=2$
$\therefore t=2$
$\Rightarrow \mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$
Since, $\sum_{k=1}^{10} f(a+k)=16\left(2^{10}-1\right)$
Then, $\sum_{k=1}^{10} 2^{a+k}=16\left(2^{10}-1\right)$
$\Rightarrow 2^{a} \sum_{k=1}^{10} 2^{k}=16\left(2^{10}-1\right)$
$\Rightarrow 2^{a} \times \frac{\left(\left(2^{10}\right)-1\right) \times 2}{(2-1)}=16 \times\left(2^{10}-1\right) \Rightarrow 2.2^{a}=16$
$\Rightarrow a=3$