Let Σlimits_{k = 1}^{10} {f,(a, + ,k)} , = ,16,({2^{10}}, - ,1), where function f satis

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1.Relation and Function

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Let $\sum\limits_{k = 1}^{10} {f\,(a\, + \,k)} \, = \,16\,({2^{10}}\, - \,1),$ 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 $‘ a '$ is

A

$4$

B

$16$

C

$2$

D

$3$

(JEE MAIN-2019)

Solution

From the given functional equation:

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

${2^{a + 1}} + {2^{a + 2}} + … + {2^{a + 10}} = 16\left( {{2^{10}} – 1} \right)$

${2^a}\left( {2 + {2^2} + … + {2^{10}}} \right) = 16\left( {{2^{10}} – 1} \right)$

${2^a}.\frac{{2.\left( {{2^{10}} – 1} \right)}}{1} = 16\left( {{2^{10}} – 1} \right)$

${2^{a + 1}} = 16 = {2^4}$

$a = 3$

Standard 12

Mathematics

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