Question

What is the greatest value of the positive integer n satisfying the condition $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...... + \dfrac{1}{{{2^{n - 1}}}} < 2 - \dfrac{1}{{1000}}$ ?
1. 8
2. 9
3. 10
4. 11

Answer 1

In this question we have to find the greatest value of the positive integer ‘n’ which satisfies the condition given in the question. So, first we will observe the series given in the question. The ratio between every term is the same. So, this will be a geometric series. By using the formulas of geometric series we will find the value of ‘n’.

Complete step-by-step solution:
Given: $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...... + \dfrac{1}{{{2^{n - 1}}}} < 2 - \dfrac{1}{{1000}}$
We will find the ratio between the terms given in LHS.
Ratio $= \dfrac{{{\text{second term}}}}{{{\text{first term}}}}$
$= \dfrac{{\dfrac{1}{2}}}{1}$
$= \dfrac{1}{2}$
Now we will find the ratio between the third and second term.
Ratio $= \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}}$
$= \dfrac{2}{4}$
$= \dfrac{1}{2}$
The ratio between the terms is the same. So, this is a geometric series.
In the LHS the terms are added to each other. So, we will apply the formula of the sum of terms in geometric series to find the value of ‘n’.
The formula for sum of geometric series is $\dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$ . Here, a is the first term of the series. r is the ratio of the terms and n is the number of terms.
Substituting the values in the above formula.
Sum of the series $= \dfrac{{1\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{1 - \dfrac{1}{2}}}$
By multiplying the reciprocal of denominator with numerator. We get,
$= 1 - {\left( {\dfrac{1}{2}} \right)^n} \times \dfrac{2}{{2 - 1}}$
$= 2 - 2 \times {\left( {\dfrac{1}{2}} \right)^n}$
$= 2 - {2^{1 - n}}$
This term is less than $2 - \dfrac{1}{{1000}}$ . So,
$2 - {2^{1 - n}} < 2 - \dfrac{1}{{1000}}$
$\Rightarrow {2^{1 - n}} < \dfrac{1}{{1000}}$
$\Rightarrow {2^{n - 1}} < 1000$.
So, the value of ‘n’ cannot be more than 10. As if the value of ‘n’ will be more than 10. Then, the value of LHS will not be less than 1000.
So, option (3) is the correct answer.

Note: If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a geometric progression. (GP), whereas the constant value is called the common ratio. The sum of a geometric series depends on the number of terms in it. The sum of a geometric series will be a definite value if the ratio’s absolute value is less than 1. If the numbers are approaching zero, they become insignificantly small. In this case, the sum to be calculated despite the series comprising infinite terms.