Question:
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
Correct Option: , 2
Solution:
From the given functional equation :
$f(x)=2^{x} \quad \forall x \in N$
$2^{a+1}+2^{a+2}+\ldots .+2^{a+10}=16\left(2^{10}-1\right)$
$2^{a}\left(2+2^{2}+\ldots .+2^{10}\right)=16\left(2^{10}-1\right)$
$2^{\mathrm{a}} \cdot \frac{2 \cdot\left(2^{10}-1\right)}{1}=16\left(2^{10}-1\right)$
$2^{\mathrm{a}+1}=16=2^{4}$
$a=3$
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