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Let $\sum_{k=1}^{10}f\left ( a+k \right )=16\left ( 2^{10-1} \right )$ , where the function $f$ satisfies $f\left ( x+y \right )=f\left ( x \right )f\left ( y \right )$ for all natural numbers $x,y$ and $f\left ( 1 \right )=2 \; .$ Then the natural numbers $'a'$ is :

Option 1)
$2$

Option 2)
$16$

Option 3)
$4$

Option 4)
$3$

Answers (1)

S solutionqc

$f\left ( 1 \right )=2$

and $f\left ( x+y \right )=f\left ( x \right )f\left ( y \right )$

put $x=1,y=1$

$f\left ( 2 \right )=f\left ( 1 \right )f\left ( 1 \right )=2\times 2=4=2^{2}$

$f\left ( 3 \right )=f\left ( 1 \right )f\left ( 2 \right )=2^{3}$ $\left ( x=1,y=2 \right )$

$f\left ( 4 \right )=f\left ( 1 \right )f\left ( 3 \right )=2^{4}$ $\left ( x=1,y=3 \right )$

Similarly $f\left ( 10 \right )$ = $2^{10}$

Now, $\sum_{k=1}^{10}f\left ( a+k \right )=16\left ( 2^{10}-1 \right )$

$\Rightarrow \sum_{k=1}^{10}f\left ( a \right )f\left ( k \right )=16\left ( 2^{10}-1 \right )$

$\Rightarrow f\left ( a \right )\sum_{k=1}^{10}f\left ( k \right )=16\left ( 2^{10}-1 \right )$

$\Rightarrow f\left ( a \right )\left [ f\left ( 1 \right )+f\left ( 2 \right )+.........+f\left ( 10 \right ) \right ]=16\left ( 2^{10}-1 \right )$

$\Rightarrow f\left ( a \right )\left [ 2^{2}+2^{2}+....+2^{10} \right ]=16\left [ 2^{10}-1 \right ]$

$\Rightarrow f\left ( a \right )\left [ \frac{2\left ( 2^{10}-1 \right )}{2-1} \right ]=16\left [ 2^{10}-1 \right ]$

$f\left ( a \right )=8=2^{3}$

$a=3$

Option 1)

$2$

Option 2)

$16$

Option 3)

$4$

Option 4)

$3$

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