Solve this following


Let $\sum_{\mathrm{k}=1}^{10} f(\mathrm{a}+\mathrm{k})=16\left(2^{10}-1\right)$, where the function

$f$ satisfies $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})$ for all natural numbers $\mathrm{x}, \mathrm{y}$ and $f(1)=2$. then the natural number ' $a$ ' is


  1. 4

  2. 3

  3. 16

  4. 2

Correct Option: , 2


From the given functional equation :

$f(x)=2^{x} \quad \forall x \in N$

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

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

$2^{\mathrm{a}} \cdot \frac{2 \cdot\left(2^{10}-1\right)}{1}=16\left(2^{10}-1\right)$




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