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$1+2+2^2+2^3+....+2^n>9999$

$\implies\frac{2^{n+1} -1}{2-1} >9999$ (sum of GP series)

$\implies 2^{n+1} >10000$

Taking log on both sides

$\implies ({n+1})*log_{10}2> 4*log_{10}10$

$\implies (n+1)*0.30103 > 4$ (given $log_{10}2=0.30103$)

$\implies n + 1 > \frac{4}{0.30103}$

$\implies n  +1 > 13.29$

$\implies n   > 12.29$

$\therefore$ Option $B.$ is the correct answer.
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