ISI2016-DCG-17

Question

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

• A. $12$
• B. $13$
• C. $14$
• D. None of these

Answer 1

Let:

(1) => $S = 1 + 2 + 2^{2} + ....... + 2^n$

 

Multiplying (1) by $2$, we get:

(2) => $2S = 2 + 2^{2} + ....... + 2^n + 2^{n + 1}$

 

Subracting (1) from (2):

(2) - (1) => $S = 2^{n + 1} - 1$

 

From the given data:

$S >= 9999$

i.e.

$2^{n + 1} - 1 >= 9999$

=> $2^{n + 1} >= 10000$

 

Taking $\log_{10}$ on both sides: 

$\log_{10}2^{n + 1} = \log_{10}10000$

 

By properties of $log$:

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

 

Substituting $\log_{10}2 = 0.30103$:

$(n+1)(0.30103) >= 4$

=> $n+1 >= 13.28771$

=> $n >= 12.28771$

 

Since $n$ is an integer:

$n = 13$

Option (B) is the correct answer.