Question

The least value of \[{\rm{n}}\] (a natural number), for which the sum S of the series \[1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots \] differs from \[{S_n}\] by a quantity \[ < {10^{ - 6}}\] is
A. 21
B. 20
C. 19
D. None

Answer 1

Hint: In our case, we are provided that the sum of the series \[1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots \] differs from \[{S_n}\] by a quantity \[ < {10^{ - 6}}\] and are questioned to determine the least value of n. Here n is a natural number. For that, we have to write the series in expression form using the infinite terms of a series formula for n terms and then applying the given condition and then solving accordingly will give the desired solution.

Complete step by step solution: We have been provided in the question that,
The sum S of the series \[1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots \] differs from \[{S_n}\] by a quantity \[ < {10^{ - 6}}\]
Now, we have to write the given series by using formula, we get
\[S = \dfrac{1}{{1 - \dfrac{1}{2}}} = 2\]
Now, we have to write the same for \[{S_n}\] we get
\[{S_n} = \dfrac{{1\left( {1 - \dfrac{1}{{{2^n}}}} \right)}}{{1 - \dfrac{1}{2}}}\]
On simplifying the above expression, we get
\[ = 2 - \dfrac{1}{{{2^{n - 1}}}}\]
We have been also given that
\[S - {S_n} < {10^{ - 6}}\]
Therefore, we can write the equation as below
\[\dfrac{1}{{{2^{n - 1}}}} < {10^{ - 6}}\]
Now, we have to rewrite the expression without fraction, we obtain
\[ \Rightarrow {2^{n - 1}} > {10^6}\]
Now, we have to simplify the power, we get
 \[n - 1 > 6{\log _2}10 = \dfrac{6}{{0.3010}}\]
Since, we have known that \[\dfrac{6}{{3018}} < 20\] then
\[ \Rightarrow n > 20\]
So, the value will be concluded as
\[n = 21\]
Therefore, the least value of \[{\rm{n}}\] (a natural number), for which the sum S of the series \[1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots \] differs from \[{S_n}\] by a quantity \[ < {10^{ - 6}}\] is \[n = 21\]

Option ‘A’ is correct
Note: These types of problems should be solved cautiously as it involves calculations having powers and exponents. And one should also be thorough with logarithmic properties to solve these types of problems. And also, they should have the knowledge about values of the powers in order to solve these types of problems to avoid wrong solution.