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The smallest integer $n$ for which $1+2+2^2+2^3+2^4+ \cdots +2^n$ exceeds $9999$, given that $\log_{10} 2=0.30103$, is

- $12$
- $13$
- $14$
- None of these

+1 vote

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

$\implies\frac{2^{n+1} -1}{2-1} >9999$ (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.

$\implies\frac{2^{n+1} -1}{2-1} >9999$ (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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