Let sumlimits_{k = 1}^{10} {f,(a, + ,k)} | vedclass

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Question

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^{1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

From the given functional equation:

$f\left( x \right) = {2^x}\forall x \in N$

${2^{a + 1}} + {2^{a + 2}} + ... + {2^{a + 10}} = 16\left( {{2^{10}} - 1} \right)$

${2^a}\left( {2 + {2^2} + ... + {2^{10}}} \right) = 16\left( {{2^{10}} - 1} \right)$

${2^a}.\frac{{2.\left( {{2^{10}} - 1} \right)}}{1} = 16\left( {{2^{10}} - 1} \right)$

${2^{a + 1}} = 16 = {2^4}$

$a = 3$

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

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